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David Howells108b42b2006-03-31 16:00:29 +01001 ============================
2 LINUX KERNEL MEMORY BARRIERS
3 ============================
4
5By: David Howells <dhowells@redhat.com>
David Howells90fddab2010-03-24 09:43:00 +00006 Paul E. McKenney <paulmck@linux.vnet.ibm.com>
Peter Zijlstrae7720af52016-04-26 10:22:05 -07007 Will Deacon <will.deacon@arm.com>
8 Peter Zijlstra <peterz@infradead.org>
David Howells108b42b2006-03-31 16:00:29 +01009
Peter Zijlstrae7720af52016-04-26 10:22:05 -070010==========
11DISCLAIMER
12==========
13
14This document is not a specification; it is intentionally (for the sake of
15brevity) and unintentionally (due to being human) incomplete. This document is
16meant as a guide to using the various memory barriers provided by Linux, but
Andrea Parri621df432018-02-20 15:25:07 -080017in case of any doubt (and there are many) please ask. Some doubts may be
18resolved by referring to the formal memory consistency model and related
19documentation at tools/memory-model/. Nevertheless, even this memory
20model should be viewed as the collective opinion of its maintainers rather
21than as an infallible oracle.
Peter Zijlstrae7720af52016-04-26 10:22:05 -070022
23To repeat, this document is not a specification of what Linux expects from
24hardware.
25
David Howells8d4840e2016-04-26 10:22:06 -070026The purpose of this document is twofold:
27
28 (1) to specify the minimum functionality that one can rely on for any
29 particular barrier, and
30
31 (2) to provide a guide as to how to use the barriers that are available.
32
33Note that an architecture can provide more than the minimum requirement
Stan Drozd35bdc722017-04-20 11:03:36 +020034for any particular barrier, but if the architecture provides less than
David Howells8d4840e2016-04-26 10:22:06 -070035that, that architecture is incorrect.
36
37Note also that it is possible that a barrier may be a no-op for an
38architecture because the way that arch works renders an explicit barrier
39unnecessary in that case.
40
41
Peter Zijlstrae7720af52016-04-26 10:22:05 -070042========
43CONTENTS
44========
David Howells108b42b2006-03-31 16:00:29 +010045
46 (*) Abstract memory access model.
47
48 - Device operations.
49 - Guarantees.
50
51 (*) What are memory barriers?
52
53 - Varieties of memory barrier.
54 - What may not be assumed about memory barriers?
Paul E. McKenneyf28f0862018-03-07 09:27:37 -080055 - Data dependency barriers (historical).
David Howells108b42b2006-03-31 16:00:29 +010056 - Control dependencies.
57 - SMP barrier pairing.
58 - Examples of memory barrier sequences.
David Howells670bd952006-06-10 09:54:12 -070059 - Read memory barriers vs load speculation.
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -070060 - Multicopy atomicity.
David Howells108b42b2006-03-31 16:00:29 +010061
62 (*) Explicit kernel barriers.
63
64 - Compiler barrier.
Jarek Poplawski81fc6322007-05-23 13:58:20 -070065 - CPU memory barriers.
David Howells108b42b2006-03-31 16:00:29 +010066 - MMIO write barrier.
67
68 (*) Implicit kernel memory barriers.
69
SeongJae Park166bda72016-04-12 08:52:50 -070070 - Lock acquisition functions.
David Howells108b42b2006-03-31 16:00:29 +010071 - Interrupt disabling functions.
David Howells50fa6102009-04-28 15:01:38 +010072 - Sleep and wake-up functions.
David Howells108b42b2006-03-31 16:00:29 +010073 - Miscellaneous functions.
74
SeongJae Park166bda72016-04-12 08:52:50 -070075 (*) Inter-CPU acquiring barrier effects.
David Howells108b42b2006-03-31 16:00:29 +010076
SeongJae Park166bda72016-04-12 08:52:50 -070077 - Acquires vs memory accesses.
78 - Acquires vs I/O accesses.
David Howells108b42b2006-03-31 16:00:29 +010079
80 (*) Where are memory barriers needed?
81
82 - Interprocessor interaction.
83 - Atomic operations.
84 - Accessing devices.
85 - Interrupts.
86
87 (*) Kernel I/O barrier effects.
88
89 (*) Assumed minimum execution ordering model.
90
91 (*) The effects of the cpu cache.
92
93 - Cache coherency.
94 - Cache coherency vs DMA.
95 - Cache coherency vs MMIO.
96
97 (*) The things CPUs get up to.
98
99 - And then there's the Alpha.
SeongJae Park01e1cd62016-04-12 08:52:51 -0700100 - Virtual Machine Guests.
David Howells108b42b2006-03-31 16:00:29 +0100101
David Howells90fddab2010-03-24 09:43:00 +0000102 (*) Example uses.
103
104 - Circular buffers.
105
David Howells108b42b2006-03-31 16:00:29 +0100106 (*) References.
107
108
109============================
110ABSTRACT MEMORY ACCESS MODEL
111============================
112
113Consider the following abstract model of the system:
114
115 : :
116 : :
117 : :
118 +-------+ : +--------+ : +-------+
119 | | : | | : | |
120 | | : | | : | |
121 | CPU 1 |<----->| Memory |<----->| CPU 2 |
122 | | : | | : | |
123 | | : | | : | |
124 +-------+ : +--------+ : +-------+
125 ^ : ^ : ^
126 | : | : |
127 | : | : |
128 | : v : |
129 | : +--------+ : |
130 | : | | : |
131 | : | | : |
132 +---------->| Device |<----------+
133 : | | :
134 : | | :
135 : +--------+ :
136 : :
137
138Each CPU executes a program that generates memory access operations. In the
139abstract CPU, memory operation ordering is very relaxed, and a CPU may actually
140perform the memory operations in any order it likes, provided program causality
141appears to be maintained. Similarly, the compiler may also arrange the
142instructions it emits in any order it likes, provided it doesn't affect the
143apparent operation of the program.
144
145So in the above diagram, the effects of the memory operations performed by a
146CPU are perceived by the rest of the system as the operations cross the
147interface between the CPU and rest of the system (the dotted lines).
148
149
150For example, consider the following sequence of events:
151
152 CPU 1 CPU 2
153 =============== ===============
154 { A == 1; B == 2 }
Alexey Dobriyan615cc2c2014-06-06 14:36:41 -0700155 A = 3; x = B;
156 B = 4; y = A;
David Howells108b42b2006-03-31 16:00:29 +0100157
158The set of accesses as seen by the memory system in the middle can be arranged
159in 24 different combinations:
160
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400161 STORE A=3, STORE B=4, y=LOAD A->3, x=LOAD B->4
162 STORE A=3, STORE B=4, x=LOAD B->4, y=LOAD A->3
163 STORE A=3, y=LOAD A->3, STORE B=4, x=LOAD B->4
164 STORE A=3, y=LOAD A->3, x=LOAD B->2, STORE B=4
165 STORE A=3, x=LOAD B->2, STORE B=4, y=LOAD A->3
166 STORE A=3, x=LOAD B->2, y=LOAD A->3, STORE B=4
167 STORE B=4, STORE A=3, y=LOAD A->3, x=LOAD B->4
David Howells108b42b2006-03-31 16:00:29 +0100168 STORE B=4, ...
169 ...
170
171and can thus result in four different combinations of values:
172
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400173 x == 2, y == 1
174 x == 2, y == 3
175 x == 4, y == 1
176 x == 4, y == 3
David Howells108b42b2006-03-31 16:00:29 +0100177
178
179Furthermore, the stores committed by a CPU to the memory system may not be
180perceived by the loads made by another CPU in the same order as the stores were
181committed.
182
183
184As a further example, consider this sequence of events:
185
186 CPU 1 CPU 2
187 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700188 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100189 B = 4; Q = P;
190 P = &B D = *Q;
191
192There is an obvious data dependency here, as the value loaded into D depends on
193the address retrieved from P by CPU 2. At the end of the sequence, any of the
194following results are possible:
195
196 (Q == &A) and (D == 1)
197 (Q == &B) and (D == 2)
198 (Q == &B) and (D == 4)
199
200Note that CPU 2 will never try and load C into D because the CPU will load P
201into Q before issuing the load of *Q.
202
203
204DEVICE OPERATIONS
205-----------------
206
207Some devices present their control interfaces as collections of memory
208locations, but the order in which the control registers are accessed is very
209important. For instance, imagine an ethernet card with a set of internal
210registers that are accessed through an address port register (A) and a data
211port register (D). To read internal register 5, the following code might then
212be used:
213
214 *A = 5;
215 x = *D;
216
217but this might show up as either of the following two sequences:
218
219 STORE *A = 5, x = LOAD *D
220 x = LOAD *D, STORE *A = 5
221
222the second of which will almost certainly result in a malfunction, since it set
223the address _after_ attempting to read the register.
224
225
226GUARANTEES
227----------
228
229There are some minimal guarantees that may be expected of a CPU:
230
231 (*) On any given CPU, dependent memory accesses will be issued in order, with
232 respect to itself. This means that for:
233
Paul E. McKenney40555942017-10-09 09:15:21 -0700234 Q = READ_ONCE(P); D = READ_ONCE(*Q);
David Howells108b42b2006-03-31 16:00:29 +0100235
236 the CPU will issue the following memory operations:
237
238 Q = LOAD P, D = LOAD *Q
239
Paul E. McKenney40555942017-10-09 09:15:21 -0700240 and always in that order. However, on DEC Alpha, READ_ONCE() also
241 emits a memory-barrier instruction, so that a DEC Alpha CPU will
242 instead issue the following memory operations:
243
244 Q = LOAD P, MEMORY_BARRIER, D = LOAD *Q, MEMORY_BARRIER
245
246 Whether on DEC Alpha or not, the READ_ONCE() also prevents compiler
247 mischief.
David Howells108b42b2006-03-31 16:00:29 +0100248
249 (*) Overlapping loads and stores within a particular CPU will appear to be
250 ordered within that CPU. This means that for:
251
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700252 a = READ_ONCE(*X); WRITE_ONCE(*X, b);
David Howells108b42b2006-03-31 16:00:29 +0100253
254 the CPU will only issue the following sequence of memory operations:
255
256 a = LOAD *X, STORE *X = b
257
258 And for:
259
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700260 WRITE_ONCE(*X, c); d = READ_ONCE(*X);
David Howells108b42b2006-03-31 16:00:29 +0100261
262 the CPU will only issue:
263
264 STORE *X = c, d = LOAD *X
265
Matt LaPlantefa00e7e2006-11-30 04:55:36 +0100266 (Loads and stores overlap if they are targeted at overlapping pieces of
David Howells108b42b2006-03-31 16:00:29 +0100267 memory).
268
269And there are a number of things that _must_ or _must_not_ be assumed:
270
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700271 (*) It _must_not_ be assumed that the compiler will do what you want
272 with memory references that are not protected by READ_ONCE() and
273 WRITE_ONCE(). Without them, the compiler is within its rights to
274 do all sorts of "creative" transformations, which are covered in
Paul E. McKenney895f5542016-01-06 14:23:03 -0800275 the COMPILER BARRIER section.
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800276
David Howells108b42b2006-03-31 16:00:29 +0100277 (*) It _must_not_ be assumed that independent loads and stores will be issued
278 in the order given. This means that for:
279
280 X = *A; Y = *B; *D = Z;
281
282 we may get any of the following sequences:
283
284 X = LOAD *A, Y = LOAD *B, STORE *D = Z
285 X = LOAD *A, STORE *D = Z, Y = LOAD *B
286 Y = LOAD *B, X = LOAD *A, STORE *D = Z
287 Y = LOAD *B, STORE *D = Z, X = LOAD *A
288 STORE *D = Z, X = LOAD *A, Y = LOAD *B
289 STORE *D = Z, Y = LOAD *B, X = LOAD *A
290
291 (*) It _must_ be assumed that overlapping memory accesses may be merged or
292 discarded. This means that for:
293
294 X = *A; Y = *(A + 4);
295
296 we may get any one of the following sequences:
297
298 X = LOAD *A; Y = LOAD *(A + 4);
299 Y = LOAD *(A + 4); X = LOAD *A;
300 {X, Y} = LOAD {*A, *(A + 4) };
301
302 And for:
303
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700304 *A = X; *(A + 4) = Y;
David Howells108b42b2006-03-31 16:00:29 +0100305
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700306 we may get any of:
David Howells108b42b2006-03-31 16:00:29 +0100307
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700308 STORE *A = X; STORE *(A + 4) = Y;
309 STORE *(A + 4) = Y; STORE *A = X;
310 STORE {*A, *(A + 4) } = {X, Y};
David Howells108b42b2006-03-31 16:00:29 +0100311
Paul E. McKenney432fbf32014-09-04 17:12:49 -0700312And there are anti-guarantees:
313
314 (*) These guarantees do not apply to bitfields, because compilers often
315 generate code to modify these using non-atomic read-modify-write
316 sequences. Do not attempt to use bitfields to synchronize parallel
317 algorithms.
318
319 (*) Even in cases where bitfields are protected by locks, all fields
320 in a given bitfield must be protected by one lock. If two fields
321 in a given bitfield are protected by different locks, the compiler's
322 non-atomic read-modify-write sequences can cause an update to one
323 field to corrupt the value of an adjacent field.
324
325 (*) These guarantees apply only to properly aligned and sized scalar
326 variables. "Properly sized" currently means variables that are
327 the same size as "char", "short", "int" and "long". "Properly
328 aligned" means the natural alignment, thus no constraints for
329 "char", two-byte alignment for "short", four-byte alignment for
330 "int", and either four-byte or eight-byte alignment for "long",
331 on 32-bit and 64-bit systems, respectively. Note that these
332 guarantees were introduced into the C11 standard, so beware when
333 using older pre-C11 compilers (for example, gcc 4.6). The portion
334 of the standard containing this guarantee is Section 3.14, which
335 defines "memory location" as follows:
336
337 memory location
338 either an object of scalar type, or a maximal sequence
339 of adjacent bit-fields all having nonzero width
340
341 NOTE 1: Two threads of execution can update and access
342 separate memory locations without interfering with
343 each other.
344
345 NOTE 2: A bit-field and an adjacent non-bit-field member
346 are in separate memory locations. The same applies
347 to two bit-fields, if one is declared inside a nested
348 structure declaration and the other is not, or if the two
349 are separated by a zero-length bit-field declaration,
350 or if they are separated by a non-bit-field member
351 declaration. It is not safe to concurrently update two
352 bit-fields in the same structure if all members declared
353 between them are also bit-fields, no matter what the
354 sizes of those intervening bit-fields happen to be.
355
David Howells108b42b2006-03-31 16:00:29 +0100356
357=========================
358WHAT ARE MEMORY BARRIERS?
359=========================
360
361As can be seen above, independent memory operations are effectively performed
362in random order, but this can be a problem for CPU-CPU interaction and for I/O.
363What is required is some way of intervening to instruct the compiler and the
364CPU to restrict the order.
365
366Memory barriers are such interventions. They impose a perceived partial
David Howells2b948952006-06-25 05:48:49 -0700367ordering over the memory operations on either side of the barrier.
368
369Such enforcement is important because the CPUs and other devices in a system
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700370can use a variety of tricks to improve performance, including reordering,
David Howells2b948952006-06-25 05:48:49 -0700371deferral and combination of memory operations; speculative loads; speculative
372branch prediction and various types of caching. Memory barriers are used to
373override or suppress these tricks, allowing the code to sanely control the
374interaction of multiple CPUs and/or devices.
David Howells108b42b2006-03-31 16:00:29 +0100375
376
377VARIETIES OF MEMORY BARRIER
378---------------------------
379
380Memory barriers come in four basic varieties:
381
382 (1) Write (or store) memory barriers.
383
384 A write memory barrier gives a guarantee that all the STORE operations
385 specified before the barrier will appear to happen before all the STORE
386 operations specified after the barrier with respect to the other
387 components of the system.
388
389 A write barrier is a partial ordering on stores only; it is not required
390 to have any effect on loads.
391
David Howells6bc39272006-06-25 05:49:22 -0700392 A CPU can be viewed as committing a sequence of store operations to the
Guilherme G. Piccoli5692fcc2017-09-21 16:29:01 -0300393 memory system as time progresses. All stores _before_ a write barrier
394 will occur _before_ all the stores after the write barrier.
David Howells108b42b2006-03-31 16:00:29 +0100395
396 [!] Note that write barriers should normally be paired with read or data
397 dependency barriers; see the "SMP barrier pairing" subsection.
398
399
400 (2) Data dependency barriers.
401
402 A data dependency barrier is a weaker form of read barrier. In the case
403 where two loads are performed such that the second depends on the result
404 of the first (eg: the first load retrieves the address to which the second
405 load will be directed), a data dependency barrier would be required to
Nikolay Borisov51de7882018-02-20 15:25:08 -0800406 make sure that the target of the second load is updated after the address
David Howells108b42b2006-03-31 16:00:29 +0100407 obtained by the first load is accessed.
408
409 A data dependency barrier is a partial ordering on interdependent loads
410 only; it is not required to have any effect on stores, independent loads
411 or overlapping loads.
412
413 As mentioned in (1), the other CPUs in the system can be viewed as
414 committing sequences of stores to the memory system that the CPU being
415 considered can then perceive. A data dependency barrier issued by the CPU
416 under consideration guarantees that for any load preceding it, if that
417 load touches one of a sequence of stores from another CPU, then by the
418 time the barrier completes, the effects of all the stores prior to that
419 touched by the load will be perceptible to any loads issued after the data
420 dependency barrier.
421
422 See the "Examples of memory barrier sequences" subsection for diagrams
423 showing the ordering constraints.
424
425 [!] Note that the first load really has to have a _data_ dependency and
426 not a control dependency. If the address for the second load is dependent
427 on the first load, but the dependency is through a conditional rather than
428 actually loading the address itself, then it's a _control_ dependency and
429 a full read barrier or better is required. See the "Control dependencies"
430 subsection for more information.
431
432 [!] Note that data dependency barriers should normally be paired with
433 write barriers; see the "SMP barrier pairing" subsection.
434
435
436 (3) Read (or load) memory barriers.
437
438 A read barrier is a data dependency barrier plus a guarantee that all the
439 LOAD operations specified before the barrier will appear to happen before
440 all the LOAD operations specified after the barrier with respect to the
441 other components of the system.
442
443 A read barrier is a partial ordering on loads only; it is not required to
444 have any effect on stores.
445
446 Read memory barriers imply data dependency barriers, and so can substitute
447 for them.
448
449 [!] Note that read barriers should normally be paired with write barriers;
450 see the "SMP barrier pairing" subsection.
451
452
453 (4) General memory barriers.
454
David Howells670bd952006-06-10 09:54:12 -0700455 A general memory barrier gives a guarantee that all the LOAD and STORE
456 operations specified before the barrier will appear to happen before all
457 the LOAD and STORE operations specified after the barrier with respect to
458 the other components of the system.
459
460 A general memory barrier is a partial ordering over both loads and stores.
David Howells108b42b2006-03-31 16:00:29 +0100461
462 General memory barriers imply both read and write memory barriers, and so
463 can substitute for either.
464
465
466And a couple of implicit varieties:
467
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100468 (5) ACQUIRE operations.
David Howells108b42b2006-03-31 16:00:29 +0100469
470 This acts as a one-way permeable barrier. It guarantees that all memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100471 operations after the ACQUIRE operation will appear to happen after the
472 ACQUIRE operation with respect to the other components of the system.
Davidlohr Bueso787df632016-04-12 08:52:55 -0700473 ACQUIRE operations include LOCK operations and both smp_load_acquire()
Andrea Parri2f359c72018-09-26 11:29:20 -0700474 and smp_cond_load_acquire() operations.
David Howells108b42b2006-03-31 16:00:29 +0100475
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100476 Memory operations that occur before an ACQUIRE operation may appear to
477 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100478
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100479 An ACQUIRE operation should almost always be paired with a RELEASE
480 operation.
David Howells108b42b2006-03-31 16:00:29 +0100481
482
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100483 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100484
485 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100486 memory operations before the RELEASE operation will appear to happen
487 before the RELEASE operation with respect to the other components of the
488 system. RELEASE operations include UNLOCK operations and
489 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100490
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100491 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100492 happen before it completes.
493
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100494 The use of ACQUIRE and RELEASE operations generally precludes the need
495 for other sorts of memory barrier (but note the exceptions mentioned in
496 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
497 pair is -not- guaranteed to act as a full memory barrier. However, after
498 an ACQUIRE on a given variable, all memory accesses preceding any prior
499 RELEASE on that same variable are guaranteed to be visible. In other
500 words, within a given variable's critical section, all accesses of all
501 previous critical sections for that variable are guaranteed to have
502 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800503
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100504 This means that ACQUIRE acts as a minimal "acquire" operation and
505 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100506
Peter Zijlstra706eeb32017-06-12 14:50:27 +0200507A subset of the atomic operations described in atomic_t.txt have ACQUIRE and
508RELEASE variants in addition to fully-ordered and relaxed (no barrier
509semantics) definitions. For compound atomics performing both a load and a
510store, ACQUIRE semantics apply only to the load and RELEASE semantics apply
511only to the store portion of the operation.
David Howells108b42b2006-03-31 16:00:29 +0100512
513Memory barriers are only required where there's a possibility of interaction
514between two CPUs or between a CPU and a device. If it can be guaranteed that
515there won't be any such interaction in any particular piece of code, then
516memory barriers are unnecessary in that piece of code.
517
518
519Note that these are the _minimum_ guarantees. Different architectures may give
520more substantial guarantees, but they may _not_ be relied upon outside of arch
521specific code.
522
523
524WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
525----------------------------------------------
526
527There are certain things that the Linux kernel memory barriers do not guarantee:
528
529 (*) There is no guarantee that any of the memory accesses specified before a
530 memory barrier will be _complete_ by the completion of a memory barrier
531 instruction; the barrier can be considered to draw a line in that CPU's
532 access queue that accesses of the appropriate type may not cross.
533
534 (*) There is no guarantee that issuing a memory barrier on one CPU will have
535 any direct effect on another CPU or any other hardware in the system. The
536 indirect effect will be the order in which the second CPU sees the effects
537 of the first CPU's accesses occur, but see the next point:
538
David Howells6bc39272006-06-25 05:49:22 -0700539 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100540 from a second CPU's accesses, even _if_ the second CPU uses a memory
541 barrier, unless the first CPU _also_ uses a matching memory barrier (see
542 the subsection on "SMP Barrier Pairing").
543
544 (*) There is no guarantee that some intervening piece of off-the-CPU
545 hardware[*] will not reorder the memory accesses. CPU cache coherency
546 mechanisms should propagate the indirect effects of a memory barrier
547 between CPUs, but might not do so in order.
548
549 [*] For information on bus mastering DMA and coherency please read:
550
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700551 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200552 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100553 Documentation/DMA-API.txt
554
555
Paul E. McKenneyf28f0862018-03-07 09:27:37 -0800556DATA DEPENDENCY BARRIERS (HISTORICAL)
557-------------------------------------
558
559As of v4.15 of the Linux kernel, an smp_read_barrier_depends() was
560added to READ_ONCE(), which means that about the only people who
561need to pay attention to this section are those working on DEC Alpha
562architecture-specific code and those working on READ_ONCE() itself.
563For those who need it, and for those who are interested in the history,
564here is the story of data-dependency barriers.
David Howells108b42b2006-03-31 16:00:29 +0100565
566The usage requirements of data dependency barriers are a little subtle, and
567it's not always obvious that they're needed. To illustrate, consider the
568following sequence of events:
569
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800570 CPU 1 CPU 2
571 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700572 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100573 B = 4;
574 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700575 WRITE_ONCE(P, &B)
576 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800577 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100578
579There's a clear data dependency here, and it would seem that by the end of the
580sequence, Q must be either &A or &B, and that:
581
582 (Q == &A) implies (D == 1)
583 (Q == &B) implies (D == 4)
584
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700585But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100586leading to the following situation:
587
588 (Q == &B) and (D == 2) ????
589
Will Deacon806654a2018-11-19 11:02:45 +0000590While this may seem like a failure of coherency or causality maintenance, it
David Howells108b42b2006-03-31 16:00:29 +0100591isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
592Alpha).
593
David Howells2b948952006-06-25 05:48:49 -0700594To deal with this, a data dependency barrier or better must be inserted
595between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100596
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800597 CPU 1 CPU 2
598 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700599 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100600 B = 4;
601 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700602 WRITE_ONCE(P, &B);
603 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800604 <data dependency barrier>
605 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100606
607This enforces the occurrence of one of the two implications, and prevents the
608third possibility from arising.
609
Paul E. McKenney92a84dd2016-01-14 14:17:04 -0800610
David Howells108b42b2006-03-31 16:00:29 +0100611[!] Note that this extremely counterintuitive situation arises most easily on
612machines with split caches, so that, for example, one cache bank processes
613even-numbered cache lines and the other bank processes odd-numbered cache
614lines. The pointer P might be stored in an odd-numbered cache line, and the
615variable B might be stored in an even-numbered cache line. Then, if the
616even-numbered bank of the reading CPU's cache is extremely busy while the
617odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700618but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100619
620
Paul E. McKenney66ce3a42017-06-30 16:18:28 -0700621A data-dependency barrier is not required to order dependent writes
622because the CPUs that the Linux kernel supports don't do writes
623until they are certain (1) that the write will actually happen, (2)
624of the location of the write, and (3) of the value to be written.
625But please carefully read the "CONTROL DEPENDENCIES" section and the
626Documentation/RCU/rcu_dereference.txt file: The compiler can and does
627break dependencies in a great many highly creative ways.
628
629 CPU 1 CPU 2
630 =============== ===============
631 { A == 1, B == 2, C = 3, P == &A, Q == &C }
632 B = 4;
633 <write barrier>
634 WRITE_ONCE(P, &B);
635 Q = READ_ONCE(P);
636 WRITE_ONCE(*Q, 5);
637
638Therefore, no data-dependency barrier is required to order the read into
639Q with the store into *Q. In other words, this outcome is prohibited,
640even without a data-dependency barrier:
641
642 (Q == &B) && (B == 4)
643
644Please note that this pattern should be rare. After all, the whole point
645of dependency ordering is to -prevent- writes to the data structure, along
646with the expensive cache misses associated with those writes. This pattern
647can be used to record rare error conditions and the like, and the CPUs'
648naturally occurring ordering prevents such records from being lost.
649
650
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700651Note well that the ordering provided by a data dependency is local to
652the CPU containing it. See the section on "Multicopy atomicity" for
653more information.
654
655
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800656The data dependency barrier is very important to the RCU system,
657for example. See rcu_assign_pointer() and rcu_dereference() in
658include/linux/rcupdate.h. This permits the current target of an RCU'd
659pointer to be replaced with a new modified target, without the replacement
660target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100661
662See also the subsection on "Cache Coherency" for a more thorough example.
663
664
665CONTROL DEPENDENCIES
666--------------------
667
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800668Control dependencies can be a bit tricky because current compilers do
669not understand them. The purpose of this section is to help you prevent
670the compiler's ignorance from breaking your code.
671
Paul E. McKenneyff382812015-02-17 10:00:06 -0800672A load-load control dependency requires a full read memory barrier, not
673simply a data dependency barrier to make it work correctly. Consider the
674following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100675
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700676 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800677 if (q) {
678 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700679 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700680 }
David Howells108b42b2006-03-31 16:00:29 +0100681
682This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800683dependency, but rather a control dependency that the CPU may short-circuit
684by attempting to predict the outcome in advance, so that other CPUs see
685the load from b as having happened before the load from a. In such a
686case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100687
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700688 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800689 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700690 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700691 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700692 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800693
694However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800695for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800696
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800697 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700698 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800699 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800700 }
701
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800702Control dependencies pair normally with other types of barriers.
703That said, please note that neither READ_ONCE() nor WRITE_ONCE()
704are optional! Without the READ_ONCE(), the compiler might combine the
705load from 'a' with other loads from 'a'. Without the WRITE_ONCE(),
706the compiler might combine the store to 'b' with other stores to 'b'.
707Either can result in highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800708
709Worse yet, if the compiler is able to prove (say) that the value of
710variable 'a' is always non-zero, it would be well within its rights
711to optimize the original example by eliminating the "if" statement
712as follows:
713
714 q = a;
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800715 b = 1; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800716
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800717So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700718
719It is tempting to try to enforce ordering on identical stores on both
720branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800721
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800722 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800723 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800724 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800725 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800726 do_something();
727 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800728 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800729 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800730 do_something_else();
731 }
732
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700733Unfortunately, current compilers will transform this as follows at high
734optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800735
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800736 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700737 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800738 WRITE_ONCE(b, 1); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800739 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800740 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800741 do_something();
742 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800743 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800744 do_something_else();
745 }
746
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700747Now there is no conditional between the load from 'a' and the store to
748'b', which means that the CPU is within its rights to reorder them:
749The conditional is absolutely required, and must be present in the
750assembly code even after all compiler optimizations have been applied.
751Therefore, if you need ordering in this example, you need explicit
752memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800753
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700754 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700755 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800756 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800757 do_something();
758 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800759 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800760 do_something_else();
761 }
762
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700763In contrast, without explicit memory barriers, two-legged-if control
764ordering is guaranteed only when the stores differ, for example:
765
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800766 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700767 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800768 WRITE_ONCE(b, 1);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700769 do_something();
770 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800771 WRITE_ONCE(b, 2);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700772 do_something_else();
773 }
774
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800775The initial READ_ONCE() is still required to prevent the compiler from
776proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800777
778In addition, you need to be careful what you do with the local variable 'q',
779otherwise the compiler might be able to guess the value and again remove
780the needed conditional. For example:
781
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800782 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800783 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800784 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800785 do_something();
786 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800787 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800788 do_something_else();
789 }
790
791If MAX is defined to be 1, then the compiler knows that (q % MAX) is
792equal to zero, in which case the compiler is within its rights to
793transform the above code into the following:
794
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800795 q = READ_ONCE(a);
pierre Kuob26cfc42017-04-07 14:37:36 +0800796 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800797 do_something_else();
798
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700799Given this transformation, the CPU is not required to respect the ordering
800between the load from variable 'a' and the store to variable 'b'. It is
801tempting to add a barrier(), but this does not help. The conditional
802is gone, and the barrier won't bring it back. Therefore, if you are
803relying on this ordering, you should make sure that MAX is greater than
804one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800805
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800806 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800807 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
808 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800809 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800810 do_something();
811 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800812 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800813 do_something_else();
814 }
815
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700816Please note once again that the stores to 'b' differ. If they were
817identical, as noted earlier, the compiler could pull this store outside
818of the 'if' statement.
819
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700820You must also be careful not to rely too much on boolean short-circuit
821evaluation. Consider this example:
822
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800823 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700824 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700825 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700826
Paul E. McKenney5af46922015-04-25 12:48:29 -0700827Because the first condition cannot fault and the second condition is
828always true, the compiler can transform this example as following,
829defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700830
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800831 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700832 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700833
834This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700835out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700836the compiler to actually emit code for a given load, it does not force
837the compiler to use the results.
838
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700839In addition, control dependencies apply only to the then-clause and
840else-clause of the if-statement in question. In particular, it does
841not necessarily apply to code following the if-statement:
842
843 q = READ_ONCE(a);
844 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800845 WRITE_ONCE(b, 1);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700846 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800847 WRITE_ONCE(b, 2);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700848 }
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800849 WRITE_ONCE(c, 1); /* BUG: No ordering against the read from 'a'. */
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700850
851It is tempting to argue that there in fact is ordering because the
852compiler cannot reorder volatile accesses and also cannot reorder
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800853the writes to 'b' with the condition. Unfortunately for this line
854of reasoning, the compiler might compile the two writes to 'b' as
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700855conditional-move instructions, as in this fanciful pseudo-assembly
856language:
857
858 ld r1,a
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700859 cmp r1,$0
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800860 cmov,ne r4,$1
861 cmov,eq r4,$2
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700862 st r4,b
863 st $1,c
864
865A weakly ordered CPU would have no dependency of any sort between the load
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800866from 'a' and the store to 'c'. The control dependencies would extend
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700867only to the pair of cmov instructions and the store depending on them.
868In short, control dependencies apply only to the stores in the then-clause
869and else-clause of the if-statement in question (including functions
870invoked by those two clauses), not to code following that if-statement.
871
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800872
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700873Note well that the ordering provided by a control dependency is local
874to the CPU containing it. See the section on "Multicopy atomicity"
875for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800876
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800877
878In summary:
879
880 (*) Control dependencies can order prior loads against later stores.
881 However, they do -not- guarantee any other sort of ordering:
882 Not prior loads against later loads, nor prior stores against
883 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800884 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800885 later loads, smp_mb().
886
Paul E. McKenney7817b792015-12-29 16:23:18 -0800887 (*) If both legs of the "if" statement begin with identical stores to
888 the same variable, then those stores must be ordered, either by
889 preceding both of them with smp_mb() or by using smp_store_release()
890 to carry out the stores. Please note that it is -not- sufficient
Paul E. McKenneya5052652016-04-12 08:52:49 -0700891 to use barrier() at beginning of each leg of the "if" statement
892 because, as shown by the example above, optimizing compilers can
893 destroy the control dependency while respecting the letter of the
894 barrier() law.
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800895
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800896 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800897 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700898 conditional must involve the prior load. If the compiler is able
899 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800900 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
901 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800902
903 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800904 dependency into nonexistence. Careful use of READ_ONCE() or
905 atomic{,64}_read() can help to preserve your control dependency.
Paul E. McKenney895f5542016-01-06 14:23:03 -0800906 Please see the COMPILER BARRIER section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800907
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700908 (*) Control dependencies apply only to the then-clause and else-clause
909 of the if-statement containing the control dependency, including
910 any functions that these two clauses call. Control dependencies
911 do -not- apply to code following the if-statement containing the
912 control dependency.
913
Paul E. McKenneyff382812015-02-17 10:00:06 -0800914 (*) Control dependencies pair normally with other types of barriers.
915
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700916 (*) Control dependencies do -not- provide multicopy atomicity. If you
917 need all the CPUs to see a given store at the same time, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100918
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800919 (*) Compilers do not understand control dependencies. It is therefore
920 your job to ensure that they do not break your code.
921
David Howells108b42b2006-03-31 16:00:29 +0100922
923SMP BARRIER PAIRING
924-------------------
925
926When dealing with CPU-CPU interactions, certain types of memory barrier should
927always be paired. A lack of appropriate pairing is almost certainly an error.
928
Paul E. McKenneyff382812015-02-17 10:00:06 -0800929General barriers pair with each other, though they also pair with most
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700930other types of barriers, albeit without multicopy atomicity. An acquire
931barrier pairs with a release barrier, but both may also pair with other
932barriers, including of course general barriers. A write barrier pairs
933with a data dependency barrier, a control dependency, an acquire barrier,
934a release barrier, a read barrier, or a general barrier. Similarly a
935read barrier, control dependency, or a data dependency barrier pairs
936with a write barrier, an acquire barrier, a release barrier, or a
937general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100938
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800939 CPU 1 CPU 2
940 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700941 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100942 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700943 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800944 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700945 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100946
947Or:
948
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800949 CPU 1 CPU 2
950 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100951 a = 1;
952 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700953 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800954 <data dependency barrier>
955 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100956
Paul E. McKenneyff382812015-02-17 10:00:06 -0800957Or even:
958
959 CPU 1 CPU 2
960 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700961 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800962 <general barrier>
Scott Tsaid92f8422017-09-20 02:16:00 +0800963 WRITE_ONCE(x, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800964 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700965 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800966 }
967
968 assert(r1 == 0 || r2 == 0);
969
David Howells108b42b2006-03-31 16:00:29 +0100970Basically, the read barrier always has to be there, even though it can be of
971the "weaker" type.
972
David Howells670bd952006-06-10 09:54:12 -0700973[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700974match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700975versa:
976
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800977 CPU 1 CPU 2
978 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700979 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
980 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800981 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700982 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
983 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700984
David Howells108b42b2006-03-31 16:00:29 +0100985
986EXAMPLES OF MEMORY BARRIER SEQUENCES
987------------------------------------
988
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700989Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100990Consider the following sequence of events:
991
992 CPU 1
993 =======================
994 STORE A = 1
995 STORE B = 2
996 STORE C = 3
997 <write barrier>
998 STORE D = 4
999 STORE E = 5
1000
1001This sequence of events is committed to the memory coherence system in an order
1002that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +02001003STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +01001004}:
1005
1006 +-------+ : :
1007 | | +------+
1008 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001009 | | : +------+ }----- \ -----> Events perceptible to
1010 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +01001011 | | : +------+ }
1012 | CPU 1 | : | B=2 | }
1013 | | +------+ }
1014 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
1015 | | +------+ } requires all stores prior to the
1016 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001017 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +01001018 | |------>| D=4 | }
1019 | | +------+
1020 +-------+ : :
1021 |
David Howells670bd952006-06-10 09:54:12 -07001022 | Sequence in which stores are committed to the
1023 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +01001024 V
1025
1026
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001027Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +01001028loads. Consider the following sequence of events:
1029
1030 CPU 1 CPU 2
1031 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -07001032 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +01001033 STORE A = 1
1034 STORE B = 2
1035 <write barrier>
1036 STORE C = &B LOAD X
1037 STORE D = 4 LOAD C (gets &B)
1038 LOAD *C (reads B)
1039
1040Without intervention, CPU 2 may perceive the events on CPU 1 in some
1041effectively random order, despite the write barrier issued by CPU 1:
1042
1043 +-------+ : : : :
1044 | | +------+ +-------+ | Sequence of update
1045 | |------>| B=2 |----- --->| Y->8 | | of perception on
1046 | | : +------+ \ +-------+ | CPU 2
1047 | CPU 1 | : | A=1 | \ --->| C->&Y | V
1048 | | +------+ | +-------+
1049 | | wwwwwwwwwwwwwwww | : :
1050 | | +------+ | : :
1051 | | : | C=&B |--- | : : +-------+
1052 | | : +------+ \ | +-------+ | |
1053 | |------>| D=4 | ----------->| C->&B |------>| |
1054 | | +------+ | +-------+ | |
1055 +-------+ : : | : : | |
1056 | : : | |
1057 | : : | CPU 2 |
1058 | +-------+ | |
1059 Apparently incorrect ---> | | B->7 |------>| |
1060 perception of B (!) | +-------+ | |
1061 | : : | |
1062 | +-------+ | |
1063 The load of X holds ---> \ | X->9 |------>| |
1064 up the maintenance \ +-------+ | |
1065 of coherence of B ----->| B->2 | +-------+
1066 +-------+
1067 : :
1068
1069
1070In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +02001071(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +01001072
1073If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -07001074and the load of *C (ie: B) on CPU 2:
1075
1076 CPU 1 CPU 2
1077 ======================= =======================
1078 { B = 7; X = 9; Y = 8; C = &Y }
1079 STORE A = 1
1080 STORE B = 2
1081 <write barrier>
1082 STORE C = &B LOAD X
1083 STORE D = 4 LOAD C (gets &B)
1084 <data dependency barrier>
1085 LOAD *C (reads B)
1086
1087then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +01001088
1089 +-------+ : : : :
1090 | | +------+ +-------+
1091 | |------>| B=2 |----- --->| Y->8 |
1092 | | : +------+ \ +-------+
1093 | CPU 1 | : | A=1 | \ --->| C->&Y |
1094 | | +------+ | +-------+
1095 | | wwwwwwwwwwwwwwww | : :
1096 | | +------+ | : :
1097 | | : | C=&B |--- | : : +-------+
1098 | | : +------+ \ | +-------+ | |
1099 | |------>| D=4 | ----------->| C->&B |------>| |
1100 | | +------+ | +-------+ | |
1101 +-------+ : : | : : | |
1102 | : : | |
1103 | : : | CPU 2 |
1104 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001105 | | X->9 |------>| |
1106 | +-------+ | |
1107 Makes sure all effects ---> \ ddddddddddddddddd | |
1108 prior to the store of C \ +-------+ | |
1109 are perceptible to ----->| B->2 |------>| |
1110 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001111 : : +-------+
1112
1113
1114And thirdly, a read barrier acts as a partial order on loads. Consider the
1115following sequence of events:
1116
1117 CPU 1 CPU 2
1118 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001119 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001120 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001121 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001122 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001123 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001124 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001125
1126Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1127some effectively random order, despite the write barrier issued by CPU 1:
1128
David Howells670bd952006-06-10 09:54:12 -07001129 +-------+ : : : :
1130 | | +------+ +-------+
1131 | |------>| A=1 |------ --->| A->0 |
1132 | | +------+ \ +-------+
1133 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1134 | | +------+ | +-------+
1135 | |------>| B=2 |--- | : :
1136 | | +------+ \ | : : +-------+
1137 +-------+ : : \ | +-------+ | |
1138 ---------->| B->2 |------>| |
1139 | +-------+ | CPU 2 |
1140 | | A->0 |------>| |
1141 | +-------+ | |
1142 | : : +-------+
1143 \ : :
1144 \ +-------+
1145 ---->| A->1 |
1146 +-------+
1147 : :
David Howells108b42b2006-03-31 16:00:29 +01001148
1149
David Howells6bc39272006-06-25 05:49:22 -07001150If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001151load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001152
David Howells670bd952006-06-10 09:54:12 -07001153 CPU 1 CPU 2
1154 ======================= =======================
1155 { A = 0, B = 9 }
1156 STORE A=1
1157 <write barrier>
1158 STORE B=2
1159 LOAD B
1160 <read barrier>
1161 LOAD A
1162
1163then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
11642:
1165
1166 +-------+ : : : :
1167 | | +------+ +-------+
1168 | |------>| A=1 |------ --->| A->0 |
1169 | | +------+ \ +-------+
1170 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1171 | | +------+ | +-------+
1172 | |------>| B=2 |--- | : :
1173 | | +------+ \ | : : +-------+
1174 +-------+ : : \ | +-------+ | |
1175 ---------->| B->2 |------>| |
1176 | +-------+ | CPU 2 |
1177 | : : | |
1178 | : : | |
1179 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1180 barrier causes all effects \ +-------+ | |
1181 prior to the storage of B ---->| A->1 |------>| |
1182 to be perceptible to CPU 2 +-------+ | |
1183 : : +-------+
1184
1185
1186To illustrate this more completely, consider what could happen if the code
1187contained a load of A either side of the read barrier:
1188
1189 CPU 1 CPU 2
1190 ======================= =======================
1191 { A = 0, B = 9 }
1192 STORE A=1
1193 <write barrier>
1194 STORE B=2
1195 LOAD B
1196 LOAD A [first load of A]
1197 <read barrier>
1198 LOAD A [second load of A]
1199
1200Even though the two loads of A both occur after the load of B, they may both
1201come up with different values:
1202
1203 +-------+ : : : :
1204 | | +------+ +-------+
1205 | |------>| A=1 |------ --->| A->0 |
1206 | | +------+ \ +-------+
1207 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1208 | | +------+ | +-------+
1209 | |------>| B=2 |--- | : :
1210 | | +------+ \ | : : +-------+
1211 +-------+ : : \ | +-------+ | |
1212 ---------->| B->2 |------>| |
1213 | +-------+ | CPU 2 |
1214 | : : | |
1215 | : : | |
1216 | +-------+ | |
1217 | | A->0 |------>| 1st |
1218 | +-------+ | |
1219 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1220 barrier causes all effects \ +-------+ | |
1221 prior to the storage of B ---->| A->1 |------>| 2nd |
1222 to be perceptible to CPU 2 +-------+ | |
1223 : : +-------+
1224
1225
1226But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1227before the read barrier completes anyway:
1228
1229 +-------+ : : : :
1230 | | +------+ +-------+
1231 | |------>| A=1 |------ --->| A->0 |
1232 | | +------+ \ +-------+
1233 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1234 | | +------+ | +-------+
1235 | |------>| B=2 |--- | : :
1236 | | +------+ \ | : : +-------+
1237 +-------+ : : \ | +-------+ | |
1238 ---------->| B->2 |------>| |
1239 | +-------+ | CPU 2 |
1240 | : : | |
1241 \ : : | |
1242 \ +-------+ | |
1243 ---->| A->1 |------>| 1st |
1244 +-------+ | |
1245 rrrrrrrrrrrrrrrrr | |
1246 +-------+ | |
1247 | A->1 |------>| 2nd |
1248 +-------+ | |
1249 : : +-------+
1250
1251
1252The guarantee is that the second load will always come up with A == 1 if the
1253load of B came up with B == 2. No such guarantee exists for the first load of
1254A; that may come up with either A == 0 or A == 1.
1255
1256
1257READ MEMORY BARRIERS VS LOAD SPECULATION
1258----------------------------------------
1259
1260Many CPUs speculate with loads: that is they see that they will need to load an
1261item from memory, and they find a time where they're not using the bus for any
1262other loads, and so do the load in advance - even though they haven't actually
1263got to that point in the instruction execution flow yet. This permits the
1264actual load instruction to potentially complete immediately because the CPU
1265already has the value to hand.
1266
1267It may turn out that the CPU didn't actually need the value - perhaps because a
1268branch circumvented the load - in which case it can discard the value or just
1269cache it for later use.
1270
1271Consider:
1272
Ingo Molnare0edc782013-11-22 11:24:53 +01001273 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001274 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001275 LOAD B
1276 DIVIDE } Divide instructions generally
1277 DIVIDE } take a long time to perform
1278 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001279
1280Which might appear as this:
1281
1282 : : +-------+
1283 +-------+ | |
1284 --->| B->2 |------>| |
1285 +-------+ | CPU 2 |
1286 : :DIVIDE | |
1287 +-------+ | |
1288 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1289 division speculates on the +-------+ ~ | |
1290 LOAD of A : : ~ | |
1291 : :DIVIDE | |
1292 : : ~ | |
1293 Once the divisions are complete --> : : ~-->| |
1294 the CPU can then perform the : : | |
1295 LOAD with immediate effect : : +-------+
1296
1297
1298Placing a read barrier or a data dependency barrier just before the second
1299load:
1300
Ingo Molnare0edc782013-11-22 11:24:53 +01001301 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001302 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001303 LOAD B
1304 DIVIDE
1305 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001306 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001307 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001308
1309will force any value speculatively obtained to be reconsidered to an extent
1310dependent on the type of barrier used. If there was no change made to the
1311speculated memory location, then the speculated value will just be used:
1312
1313 : : +-------+
1314 +-------+ | |
1315 --->| B->2 |------>| |
1316 +-------+ | CPU 2 |
1317 : :DIVIDE | |
1318 +-------+ | |
1319 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1320 division speculates on the +-------+ ~ | |
1321 LOAD of A : : ~ | |
1322 : :DIVIDE | |
1323 : : ~ | |
1324 : : ~ | |
1325 rrrrrrrrrrrrrrrr~ | |
1326 : : ~ | |
1327 : : ~-->| |
1328 : : | |
1329 : : +-------+
1330
1331
1332but if there was an update or an invalidation from another CPU pending, then
1333the speculation will be cancelled and the value reloaded:
1334
1335 : : +-------+
1336 +-------+ | |
1337 --->| B->2 |------>| |
1338 +-------+ | CPU 2 |
1339 : :DIVIDE | |
1340 +-------+ | |
1341 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1342 division speculates on the +-------+ ~ | |
1343 LOAD of A : : ~ | |
1344 : :DIVIDE | |
1345 : : ~ | |
1346 : : ~ | |
1347 rrrrrrrrrrrrrrrrr | |
1348 +-------+ | |
1349 The speculation is discarded ---> --->| A->1 |------>| |
1350 and an updated value is +-------+ | |
1351 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001352
1353
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001354MULTICOPY ATOMICITY
1355--------------------
Paul E. McKenney241e6662011-02-10 16:54:50 -08001356
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001357Multicopy atomicity is a deeply intuitive notion about ordering that is
1358not always provided by real computer systems, namely that a given store
Alan Stern0902b1f2017-09-01 07:53:34 -07001359becomes visible at the same time to all CPUs, or, alternatively, that all
1360CPUs agree on the order in which all stores become visible. However,
1361support of full multicopy atomicity would rule out valuable hardware
1362optimizations, so a weaker form called ``other multicopy atomicity''
1363instead guarantees only that a given store becomes visible at the same
1364time to all -other- CPUs. The remainder of this document discusses this
1365weaker form, but for brevity will call it simply ``multicopy atomicity''.
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001366
1367The following example demonstrates multicopy atomicity:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001368
1369 CPU 1 CPU 2 CPU 3
1370 ======================= ======================= =======================
1371 { X = 0, Y = 0 }
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001372 STORE X=1 r1=LOAD X (reads 1) LOAD Y (reads 1)
1373 <general barrier> <read barrier>
1374 STORE Y=r1 LOAD X
Paul E. McKenney241e6662011-02-10 16:54:50 -08001375
Alan Stern0902b1f2017-09-01 07:53:34 -07001376Suppose that CPU 2's load from X returns 1, which it then stores to Y,
1377and CPU 3's load from Y returns 1. This indicates that CPU 1's store
1378to X precedes CPU 2's load from X and that CPU 2's store to Y precedes
1379CPU 3's load from Y. In addition, the memory barriers guarantee that
1380CPU 2 executes its load before its store, and CPU 3 loads from Y before
1381it loads from X. The question is then "Can CPU 3's load from X return 0?"
Paul E. McKenney241e6662011-02-10 16:54:50 -08001382
Alan Stern0902b1f2017-09-01 07:53:34 -07001383Because CPU 3's load from X in some sense comes after CPU 2's load, it
Paul E. McKenney241e6662011-02-10 16:54:50 -08001384is natural to expect that CPU 3's load from X must therefore return 1.
Alan Stern0902b1f2017-09-01 07:53:34 -07001385This expectation follows from multicopy atomicity: if a load executing
1386on CPU B follows a load from the same variable executing on CPU A (and
1387CPU A did not originally store the value which it read), then on
1388multicopy-atomic systems, CPU B's load must return either the same value
1389that CPU A's load did or some later value. However, the Linux kernel
1390does not require systems to be multicopy atomic.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001391
Alan Stern0902b1f2017-09-01 07:53:34 -07001392The use of a general memory barrier in the example above compensates
1393for any lack of multicopy atomicity. In the example, if CPU 2's load
1394from X returns 1 and CPU 3's load from Y returns 1, then CPU 3's load
1395from X must indeed also return 1.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001396
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001397However, dependencies, read barriers, and write barriers are not always
1398able to compensate for non-multicopy atomicity. For example, suppose
1399that CPU 2's general barrier is removed from the above example, leaving
1400only the data dependency shown below:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001401
1402 CPU 1 CPU 2 CPU 3
1403 ======================= ======================= =======================
1404 { X = 0, Y = 0 }
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001405 STORE X=1 r1=LOAD X (reads 1) LOAD Y (reads 1)
1406 <data dependency> <read barrier>
1407 STORE Y=r1 LOAD X (reads 0)
Paul E. McKenney241e6662011-02-10 16:54:50 -08001408
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001409This substitution allows non-multicopy atomicity to run rampant: in
1410this example, it is perfectly legal for CPU 2's load from X to return 1,
1411CPU 3's load from Y to return 1, and its load from X to return 0.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001412
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001413The key point is that although CPU 2's data dependency orders its load
Alan Stern0902b1f2017-09-01 07:53:34 -07001414and store, it does not guarantee to order CPU 1's store. Thus, if this
1415example runs on a non-multicopy-atomic system where CPUs 1 and 2 share a
1416store buffer or a level of cache, CPU 2 might have early access to CPU 1's
1417writes. General barriers are therefore required to ensure that all CPUs
1418agree on the combined order of multiple accesses.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001419
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001420General barriers can compensate not only for non-multicopy atomicity,
1421but can also generate additional ordering that can ensure that -all-
1422CPUs will perceive the same order of -all- operations. In contrast, a
1423chain of release-acquire pairs do not provide this additional ordering,
1424which means that only those CPUs on the chain are guaranteed to agree
1425on the combined order of the accesses. For example, switching to C code
1426in deference to the ghost of Herman Hollerith:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001427
1428 int u, v, x, y, z;
1429
1430 void cpu0(void)
1431 {
1432 r0 = smp_load_acquire(&x);
1433 WRITE_ONCE(u, 1);
1434 smp_store_release(&y, 1);
1435 }
1436
1437 void cpu1(void)
1438 {
1439 r1 = smp_load_acquire(&y);
1440 r4 = READ_ONCE(v);
1441 r5 = READ_ONCE(u);
1442 smp_store_release(&z, 1);
1443 }
1444
1445 void cpu2(void)
1446 {
1447 r2 = smp_load_acquire(&z);
1448 smp_store_release(&x, 1);
1449 }
1450
1451 void cpu3(void)
1452 {
1453 WRITE_ONCE(v, 1);
1454 smp_mb();
1455 r3 = READ_ONCE(u);
1456 }
1457
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001458Because cpu0(), cpu1(), and cpu2() participate in a chain of
1459smp_store_release()/smp_load_acquire() pairs, the following outcome
1460is prohibited:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001461
1462 r0 == 1 && r1 == 1 && r2 == 1
1463
1464Furthermore, because of the release-acquire relationship between cpu0()
1465and cpu1(), cpu1() must see cpu0()'s writes, so that the following
1466outcome is prohibited:
1467
1468 r1 == 1 && r5 == 0
1469
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001470However, the ordering provided by a release-acquire chain is local
1471to the CPUs participating in that chain and does not apply to cpu3(),
1472at least aside from stores. Therefore, the following outcome is possible:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001473
1474 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
1475
Paul E. McKenney37ef0342016-01-25 22:12:34 -08001476As an aside, the following outcome is also possible:
1477
1478 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
1479
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001480Although cpu0(), cpu1(), and cpu2() will see their respective reads and
1481writes in order, CPUs not involved in the release-acquire chain might
1482well disagree on the order. This disagreement stems from the fact that
1483the weak memory-barrier instructions used to implement smp_load_acquire()
1484and smp_store_release() are not required to order prior stores against
1485subsequent loads in all cases. This means that cpu3() can see cpu0()'s
1486store to u as happening -after- cpu1()'s load from v, even though
1487both cpu0() and cpu1() agree that these two operations occurred in the
1488intended order.
1489
1490However, please keep in mind that smp_load_acquire() is not magic.
1491In particular, it simply reads from its argument with ordering. It does
1492-not- ensure that any particular value will be read. Therefore, the
1493following outcome is possible:
1494
1495 r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
1496
1497Note that this outcome can happen even on a mythical sequentially
1498consistent system where nothing is ever reordered.
1499
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001500To reiterate, if your code requires full ordering of all operations,
1501use general barriers throughout.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001502
1503
David Howells108b42b2006-03-31 16:00:29 +01001504========================
1505EXPLICIT KERNEL BARRIERS
1506========================
1507
1508The Linux kernel has a variety of different barriers that act at different
1509levels:
1510
1511 (*) Compiler barrier.
1512
1513 (*) CPU memory barriers.
1514
1515 (*) MMIO write barrier.
1516
1517
1518COMPILER BARRIER
1519----------------
1520
1521The Linux kernel has an explicit compiler barrier function that prevents the
1522compiler from moving the memory accesses either side of it to the other side:
1523
1524 barrier();
1525
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001526This is a general barrier -- there are no read-read or write-write
1527variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1528thought of as weak forms of barrier() that affect only the specific
1529accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001530
Paul E. McKenney692118d2013-12-11 13:59:07 -08001531The barrier() function has the following effects:
1532
1533 (*) Prevents the compiler from reordering accesses following the
1534 barrier() to precede any accesses preceding the barrier().
1535 One example use for this property is to ease communication between
1536 interrupt-handler code and the code that was interrupted.
1537
1538 (*) Within a loop, forces the compiler to load the variables used
1539 in that loop's conditional on each pass through that loop.
1540
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001541The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1542optimizations that, while perfectly safe in single-threaded code, can
1543be fatal in concurrent code. Here are some examples of these sorts
1544of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001545
Paul E. McKenney449f7412014-01-02 15:03:50 -08001546 (*) The compiler is within its rights to reorder loads and stores
1547 to the same variable, and in some cases, the CPU is within its
1548 rights to reorder loads to the same variable. This means that
1549 the following code:
1550
1551 a[0] = x;
1552 a[1] = x;
1553
1554 Might result in an older value of x stored in a[1] than in a[0].
1555 Prevent both the compiler and the CPU from doing this as follows:
1556
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001557 a[0] = READ_ONCE(x);
1558 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001559
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001560 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1561 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001562
Paul E. McKenney692118d2013-12-11 13:59:07 -08001563 (*) The compiler is within its rights to merge successive loads from
1564 the same variable. Such merging can cause the compiler to "optimize"
1565 the following code:
1566
1567 while (tmp = a)
1568 do_something_with(tmp);
1569
1570 into the following code, which, although in some sense legitimate
1571 for single-threaded code, is almost certainly not what the developer
1572 intended:
1573
1574 if (tmp = a)
1575 for (;;)
1576 do_something_with(tmp);
1577
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001578 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001579
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001580 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001581 do_something_with(tmp);
1582
1583 (*) The compiler is within its rights to reload a variable, for example,
1584 in cases where high register pressure prevents the compiler from
1585 keeping all data of interest in registers. The compiler might
1586 therefore optimize the variable 'tmp' out of our previous example:
1587
1588 while (tmp = a)
1589 do_something_with(tmp);
1590
1591 This could result in the following code, which is perfectly safe in
1592 single-threaded code, but can be fatal in concurrent code:
1593
1594 while (a)
1595 do_something_with(a);
1596
1597 For example, the optimized version of this code could result in
1598 passing a zero to do_something_with() in the case where the variable
1599 a was modified by some other CPU between the "while" statement and
1600 the call to do_something_with().
1601
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001602 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001603
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001604 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001605 do_something_with(tmp);
1606
1607 Note that if the compiler runs short of registers, it might save
1608 tmp onto the stack. The overhead of this saving and later restoring
1609 is why compilers reload variables. Doing so is perfectly safe for
1610 single-threaded code, so you need to tell the compiler about cases
1611 where it is not safe.
1612
1613 (*) The compiler is within its rights to omit a load entirely if it knows
1614 what the value will be. For example, if the compiler can prove that
1615 the value of variable 'a' is always zero, it can optimize this code:
1616
1617 while (tmp = a)
1618 do_something_with(tmp);
1619
1620 Into this:
1621
1622 do { } while (0);
1623
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001624 This transformation is a win for single-threaded code because it
1625 gets rid of a load and a branch. The problem is that the compiler
1626 will carry out its proof assuming that the current CPU is the only
1627 one updating variable 'a'. If variable 'a' is shared, then the
1628 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1629 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001630
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001631 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001632 do_something_with(tmp);
1633
1634 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001635 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001636 do the following and MAX is a preprocessor macro with the value 1:
1637
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001638 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001639 do_something_with(tmp);
1640
1641 Then the compiler knows that the result of the "%" operator applied
1642 to MAX will always be zero, again allowing the compiler to optimize
1643 the code into near-nonexistence. (It will still load from the
1644 variable 'a'.)
1645
1646 (*) Similarly, the compiler is within its rights to omit a store entirely
1647 if it knows that the variable already has the value being stored.
1648 Again, the compiler assumes that the current CPU is the only one
1649 storing into the variable, which can cause the compiler to do the
1650 wrong thing for shared variables. For example, suppose you have
1651 the following:
1652
1653 a = 0;
SeongJae Park65f95ff2016-02-22 08:28:29 -08001654 ... Code that does not store to variable a ...
Paul E. McKenney692118d2013-12-11 13:59:07 -08001655 a = 0;
1656
1657 The compiler sees that the value of variable 'a' is already zero, so
1658 it might well omit the second store. This would come as a fatal
1659 surprise if some other CPU might have stored to variable 'a' in the
1660 meantime.
1661
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001662 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001663 wrong guess:
1664
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001665 WRITE_ONCE(a, 0);
SeongJae Park65f95ff2016-02-22 08:28:29 -08001666 ... Code that does not store to variable a ...
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001667 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001668
1669 (*) The compiler is within its rights to reorder memory accesses unless
1670 you tell it not to. For example, consider the following interaction
1671 between process-level code and an interrupt handler:
1672
1673 void process_level(void)
1674 {
1675 msg = get_message();
1676 flag = true;
1677 }
1678
1679 void interrupt_handler(void)
1680 {
1681 if (flag)
1682 process_message(msg);
1683 }
1684
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001685 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001686 process_level() to the following, in fact, this might well be a
1687 win for single-threaded code:
1688
1689 void process_level(void)
1690 {
1691 flag = true;
1692 msg = get_message();
1693 }
1694
1695 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001696 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001697 to prevent this as follows:
1698
1699 void process_level(void)
1700 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001701 WRITE_ONCE(msg, get_message());
1702 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001703 }
1704
1705 void interrupt_handler(void)
1706 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001707 if (READ_ONCE(flag))
1708 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001709 }
1710
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001711 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1712 interrupt_handler() are needed if this interrupt handler can itself
1713 be interrupted by something that also accesses 'flag' and 'msg',
1714 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1715 and WRITE_ONCE() are not needed in interrupt_handler() other than
1716 for documentation purposes. (Note also that nested interrupts
1717 do not typically occur in modern Linux kernels, in fact, if an
1718 interrupt handler returns with interrupts enabled, you will get a
1719 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001720
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001721 You should assume that the compiler can move READ_ONCE() and
1722 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1723 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001724
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001725 This effect could also be achieved using barrier(), but READ_ONCE()
1726 and WRITE_ONCE() are more selective: With READ_ONCE() and
1727 WRITE_ONCE(), the compiler need only forget the contents of the
1728 indicated memory locations, while with barrier() the compiler must
1729 discard the value of all memory locations that it has currented
1730 cached in any machine registers. Of course, the compiler must also
1731 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1732 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001733
1734 (*) The compiler is within its rights to invent stores to a variable,
1735 as in the following example:
1736
1737 if (a)
1738 b = a;
1739 else
1740 b = 42;
1741
1742 The compiler might save a branch by optimizing this as follows:
1743
1744 b = 42;
1745 if (a)
1746 b = a;
1747
1748 In single-threaded code, this is not only safe, but also saves
1749 a branch. Unfortunately, in concurrent code, this optimization
1750 could cause some other CPU to see a spurious value of 42 -- even
1751 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001752 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001753
1754 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001755 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001756 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001757 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001758
1759 The compiler can also invent loads. These are usually less
1760 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001761 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001762 invented loads.
1763
1764 (*) For aligned memory locations whose size allows them to be accessed
1765 with a single memory-reference instruction, prevents "load tearing"
1766 and "store tearing," in which a single large access is replaced by
1767 multiple smaller accesses. For example, given an architecture having
1768 16-bit store instructions with 7-bit immediate fields, the compiler
1769 might be tempted to use two 16-bit store-immediate instructions to
1770 implement the following 32-bit store:
1771
1772 p = 0x00010002;
1773
1774 Please note that GCC really does use this sort of optimization,
1775 which is not surprising given that it would likely take more
1776 than two instructions to build the constant and then store it.
1777 This optimization can therefore be a win in single-threaded code.
1778 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1779 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001780 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001781
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001782 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001783
1784 Use of packed structures can also result in load and store tearing,
1785 as in this example:
1786
1787 struct __attribute__((__packed__)) foo {
1788 short a;
1789 int b;
1790 short c;
1791 };
1792 struct foo foo1, foo2;
1793 ...
1794
1795 foo2.a = foo1.a;
1796 foo2.b = foo1.b;
1797 foo2.c = foo1.c;
1798
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001799 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1800 volatile markings, the compiler would be well within its rights to
1801 implement these three assignment statements as a pair of 32-bit
1802 loads followed by a pair of 32-bit stores. This would result in
1803 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1804 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001805
1806 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001807 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001808 foo2.c = foo1.c;
1809
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001810All that aside, it is never necessary to use READ_ONCE() and
1811WRITE_ONCE() on a variable that has been marked volatile. For example,
1812because 'jiffies' is marked volatile, it is never necessary to
1813say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1814WRITE_ONCE() are implemented as volatile casts, which has no effect when
1815its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001816
1817Please note that these compiler barriers have no direct effect on the CPU,
1818which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001819
1820
1821CPU MEMORY BARRIERS
1822-------------------
1823
1824The Linux kernel has eight basic CPU memory barriers:
1825
1826 TYPE MANDATORY SMP CONDITIONAL
1827 =============== ======================= ===========================
1828 GENERAL mb() smp_mb()
1829 WRITE wmb() smp_wmb()
1830 READ rmb() smp_rmb()
Paul E. McKenney9ad3c142017-11-27 09:20:40 -08001831 DATA DEPENDENCY READ_ONCE()
David Howells108b42b2006-03-31 16:00:29 +01001832
1833
Nick Piggin73f10282008-05-14 06:35:11 +02001834All memory barriers except the data dependency barriers imply a compiler
SeongJae Park0b6fa342016-04-12 08:52:53 -07001835barrier. Data dependencies do not impose any additional compiler ordering.
Nick Piggin73f10282008-05-14 06:35:11 +02001836
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001837Aside: In the case of data dependencies, the compiler would be expected
1838to issue the loads in the correct order (eg. `a[b]` would have to load
1839the value of b before loading a[b]), however there is no guarantee in
1840the C specification that the compiler may not speculate the value of b
1841(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
SeongJae Park0b6fa342016-04-12 08:52:53 -07001842tmp = a[b]; ). There is also the problem of a compiler reloading b after
1843having loaded a[b], thus having a newer copy of b than a[b]. A consensus
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001844has not yet been reached about these problems, however the READ_ONCE()
1845macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001846
1847SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001848systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001849and will order overlapping accesses correctly with respect to itself.
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001850However, see the subsection on "Virtual Machine Guests" below.
David Howells108b42b2006-03-31 16:00:29 +01001851
1852[!] Note that SMP memory barriers _must_ be used to control the ordering of
1853references to shared memory on SMP systems, though the use of locking instead
1854is sufficient.
1855
1856Mandatory barriers should not be used to control SMP effects, since mandatory
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001857barriers impose unnecessary overhead on both SMP and UP systems. They may,
1858however, be used to control MMIO effects on accesses through relaxed memory I/O
1859windows. These barriers are required even on non-SMP systems as they affect
1860the order in which memory operations appear to a device by prohibiting both the
1861compiler and the CPU from reordering them.
David Howells108b42b2006-03-31 16:00:29 +01001862
1863
1864There are some more advanced barrier functions:
1865
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001866 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001867
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001868 This assigns the value to the variable and then inserts a full memory
Davidlohr Bueso2d142e52015-10-27 12:53:51 -07001869 barrier after it. It isn't guaranteed to insert anything more than a
1870 compiler barrier in a UP compilation.
David Howells108b42b2006-03-31 16:00:29 +01001871
1872
Peter Zijlstra1b156112014-03-13 19:00:35 +01001873 (*) smp_mb__before_atomic();
1874 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001875
Peter Zijlstra1b156112014-03-13 19:00:35 +01001876 These are for use with atomic (such as add, subtract, increment and
1877 decrement) functions that don't return a value, especially when used for
1878 reference counting. These functions do not imply memory barriers.
1879
1880 These are also used for atomic bitop functions that do not return a
1881 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001882
1883 As an example, consider a piece of code that marks an object as being dead
1884 and then decrements the object's reference count:
1885
1886 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001887 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001888 atomic_dec(&obj->ref_count);
1889
1890 This makes sure that the death mark on the object is perceived to be set
1891 *before* the reference counter is decremented.
1892
Peter Zijlstra706eeb32017-06-12 14:50:27 +02001893 See Documentation/atomic_{t,bitops}.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01001894
1895
Alexander Duyck1077fa32014-12-11 15:02:06 -08001896 (*) dma_wmb();
1897 (*) dma_rmb();
1898
1899 These are for use with consistent memory to guarantee the ordering
1900 of writes or reads of shared memory accessible to both the CPU and a
1901 DMA capable device.
1902
1903 For example, consider a device driver that shares memory with a device
1904 and uses a descriptor status value to indicate if the descriptor belongs
1905 to the device or the CPU, and a doorbell to notify it when new
1906 descriptors are available:
1907
1908 if (desc->status != DEVICE_OWN) {
1909 /* do not read data until we own descriptor */
1910 dma_rmb();
1911
1912 /* read/modify data */
1913 read_data = desc->data;
1914 desc->data = write_data;
1915
1916 /* flush modifications before status update */
1917 dma_wmb();
1918
1919 /* assign ownership */
1920 desc->status = DEVICE_OWN;
1921
Alexander Duyck1077fa32014-12-11 15:02:06 -08001922 /* notify device of new descriptors */
1923 writel(DESC_NOTIFY, doorbell);
1924 }
1925
1926 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001927 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001928 us to guarantee the data is written to the descriptor before the device
Will Deacon58465812018-05-14 15:55:26 -07001929 can see it now has ownership. Note that, when using writel(), a prior
1930 wmb() is not needed to guarantee that the cache coherent memory writes
1931 have completed before writing to the MMIO region. The cheaper
1932 writel_relaxed() does not provide this guarantee and must not be used
1933 here.
Alexander Duyck1077fa32014-12-11 15:02:06 -08001934
Will Deacon58465812018-05-14 15:55:26 -07001935 See the subsection "Kernel I/O barrier effects" for more information on
1936 relaxed I/O accessors and the Documentation/DMA-API.txt file for more
1937 information on consistent memory.
Alexander Duyck1077fa32014-12-11 15:02:06 -08001938
SeongJae Parkdfeccea2016-08-11 11:17:40 -07001939
David Howells108b42b2006-03-31 16:00:29 +01001940MMIO WRITE BARRIER
1941------------------
1942
1943The Linux kernel also has a special barrier for use with memory-mapped I/O
1944writes:
1945
1946 mmiowb();
1947
1948This is a variation on the mandatory write barrier that causes writes to weakly
1949ordered I/O regions to be partially ordered. Its effects may go beyond the
1950CPU->Hardware interface and actually affect the hardware at some level.
1951
SeongJae Park166bda72016-04-12 08:52:50 -07001952See the subsection "Acquires vs I/O accesses" for more information.
David Howells108b42b2006-03-31 16:00:29 +01001953
1954
1955===============================
1956IMPLICIT KERNEL MEMORY BARRIERS
1957===============================
1958
1959Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001960which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001961
1962This specification is a _minimum_ guarantee; any particular architecture may
1963provide more substantial guarantees, but these may not be relied upon outside
1964of arch specific code.
1965
1966
SeongJae Park166bda72016-04-12 08:52:50 -07001967LOCK ACQUISITION FUNCTIONS
1968--------------------------
David Howells108b42b2006-03-31 16:00:29 +01001969
1970The Linux kernel has a number of locking constructs:
1971
1972 (*) spin locks
1973 (*) R/W spin locks
1974 (*) mutexes
1975 (*) semaphores
1976 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001977
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001978In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001979for each construct. These operations all imply certain barriers:
1980
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001981 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001982
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001983 Memory operations issued after the ACQUIRE will be completed after the
1984 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001985
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001986 Memory operations issued before the ACQUIRE may be completed after
Peter Zijlstraa9668cd2017-06-07 17:51:27 +02001987 the ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001988
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001989 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001990
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001991 Memory operations issued before the RELEASE will be completed before the
1992 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001993
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001994 Memory operations issued after the RELEASE may be completed before the
1995 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001996
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001997 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001998
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001999 All ACQUIRE operations issued before another ACQUIRE operation will be
2000 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01002001
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002002 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01002003
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002004 All ACQUIRE operations issued before a RELEASE operation will be
2005 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01002006
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002007 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01002008
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002009 Certain locking variants of the ACQUIRE operation may fail, either due to
2010 being unable to get the lock immediately, or due to receiving an unblocked
Will Deacon806654a2018-11-19 11:02:45 +00002011 signal while asleep waiting for the lock to become available. Failed
David Howells108b42b2006-03-31 16:00:29 +01002012 locks do not imply any sort of barrier.
2013
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002014[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
2015one-way barriers is that the effects of instructions outside of a critical
2016section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01002017
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002018An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
2019because it is possible for an access preceding the ACQUIRE to happen after the
2020ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
2021the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07002022
2023 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002024 ACQUIRE M
2025 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07002026 *B = b;
2027
2028may occur as:
2029
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002030 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002031
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002032When the ACQUIRE and RELEASE are a lock acquisition and release,
2033respectively, this same reordering can occur if the lock's ACQUIRE and
2034RELEASE are to the same lock variable, but only from the perspective of
2035another CPU not holding that lock. In short, a ACQUIRE followed by an
2036RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002037
Paul E. McKenney12d560f2015-07-14 18:35:23 -07002038Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
2039not imply a full memory barrier. Therefore, the CPU's execution of the
2040critical sections corresponding to the RELEASE and the ACQUIRE can cross,
2041so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002042
2043 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002044 RELEASE M
2045 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002046 *B = b;
2047
2048could occur as:
2049
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002050 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002051
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002052It might appear that this reordering could introduce a deadlock.
2053However, this cannot happen because if such a deadlock threatened,
2054the RELEASE would simply complete, thereby avoiding the deadlock.
2055
2056 Why does this work?
2057
2058 One key point is that we are only talking about the CPU doing
2059 the reordering, not the compiler. If the compiler (or, for
2060 that matter, the developer) switched the operations, deadlock
2061 -could- occur.
2062
2063 But suppose the CPU reordered the operations. In this case,
2064 the unlock precedes the lock in the assembly code. The CPU
2065 simply elected to try executing the later lock operation first.
2066 If there is a deadlock, this lock operation will simply spin (or
2067 try to sleep, but more on that later). The CPU will eventually
2068 execute the unlock operation (which preceded the lock operation
2069 in the assembly code), which will unravel the potential deadlock,
2070 allowing the lock operation to succeed.
2071
2072 But what if the lock is a sleeplock? In that case, the code will
2073 try to enter the scheduler, where it will eventually encounter
2074 a memory barrier, which will force the earlier unlock operation
2075 to complete, again unraveling the deadlock. There might be
2076 a sleep-unlock race, but the locking primitive needs to resolve
2077 such races properly in any case.
2078
David Howells108b42b2006-03-31 16:00:29 +01002079Locks and semaphores may not provide any guarantee of ordering on UP compiled
2080systems, and so cannot be counted on in such a situation to actually achieve
2081anything at all - especially with respect to I/O accesses - unless combined
2082with interrupt disabling operations.
2083
SeongJae Parkd7cab362016-08-11 11:17:41 -07002084See also the section on "Inter-CPU acquiring barrier effects".
David Howells108b42b2006-03-31 16:00:29 +01002085
2086
2087As an example, consider the following:
2088
2089 *A = a;
2090 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002091 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01002092 *C = c;
2093 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002094 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002095 *E = e;
2096 *F = f;
2097
2098The following sequence of events is acceptable:
2099
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002100 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002101
2102 [+] Note that {*F,*A} indicates a combined access.
2103
2104But none of the following are:
2105
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002106 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
2107 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
2108 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
2109 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01002110
2111
2112
2113INTERRUPT DISABLING FUNCTIONS
2114-----------------------------
2115
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002116Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
2117(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01002118barriers are required in such a situation, they must be provided from some
2119other means.
2120
2121
David Howells50fa6102009-04-28 15:01:38 +01002122SLEEP AND WAKE-UP FUNCTIONS
2123---------------------------
2124
2125Sleeping and waking on an event flagged in global data can be viewed as an
2126interaction between two pieces of data: the task state of the task waiting for
2127the event and the global data used to indicate the event. To make sure that
2128these appear to happen in the right order, the primitives to begin the process
2129of going to sleep, and the primitives to initiate a wake up imply certain
2130barriers.
2131
2132Firstly, the sleeper normally follows something like this sequence of events:
2133
2134 for (;;) {
2135 set_current_state(TASK_UNINTERRUPTIBLE);
2136 if (event_indicated)
2137 break;
2138 schedule();
2139 }
2140
2141A general memory barrier is interpolated automatically by set_current_state()
2142after it has altered the task state:
2143
2144 CPU 1
2145 ===============================
2146 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002147 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01002148 STORE current->state
2149 <general barrier>
2150 LOAD event_indicated
2151
2152set_current_state() may be wrapped by:
2153
2154 prepare_to_wait();
2155 prepare_to_wait_exclusive();
2156
2157which therefore also imply a general memory barrier after setting the state.
2158The whole sequence above is available in various canned forms, all of which
2159interpolate the memory barrier in the right place:
2160
2161 wait_event();
2162 wait_event_interruptible();
2163 wait_event_interruptible_exclusive();
2164 wait_event_interruptible_timeout();
2165 wait_event_killable();
2166 wait_event_timeout();
2167 wait_on_bit();
2168 wait_on_bit_lock();
2169
2170
2171Secondly, code that performs a wake up normally follows something like this:
2172
2173 event_indicated = 1;
2174 wake_up(&event_wait_queue);
2175
2176or:
2177
2178 event_indicated = 1;
2179 wake_up_process(event_daemon);
2180
Andrea Parri7696f992018-07-16 11:06:03 -07002181A general memory barrier is executed by wake_up() if it wakes something up.
2182If it doesn't wake anything up then a memory barrier may or may not be
2183executed; you must not rely on it. The barrier occurs before the task state
2184is accessed, in particular, it sits between the STORE to indicate the event
2185and the STORE to set TASK_RUNNING:
David Howells50fa6102009-04-28 15:01:38 +01002186
Andrea Parri7696f992018-07-16 11:06:03 -07002187 CPU 1 (Sleeper) CPU 2 (Waker)
David Howells50fa6102009-04-28 15:01:38 +01002188 =============================== ===============================
2189 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002190 smp_store_mb(); wake_up();
Andrea Parri7696f992018-07-16 11:06:03 -07002191 STORE current->state ...
2192 <general barrier> <general barrier>
2193 LOAD event_indicated if ((LOAD task->state) & TASK_NORMAL)
2194 STORE task->state
David Howells50fa6102009-04-28 15:01:38 +01002195
Andrea Parri7696f992018-07-16 11:06:03 -07002196where "task" is the thread being woken up and it equals CPU 1's "current".
2197
2198To repeat, a general memory barrier is guaranteed to be executed by wake_up()
2199if something is actually awakened, but otherwise there is no such guarantee.
2200To see this, consider the following sequence of events, where X and Y are both
2201initially zero:
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002202
2203 CPU 1 CPU 2
2204 =============================== ===============================
Andrea Parri7696f992018-07-16 11:06:03 -07002205 X = 1; Y = 1;
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002206 smp_mb(); wake_up();
Andrea Parri7696f992018-07-16 11:06:03 -07002207 LOAD Y LOAD X
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002208
Andrea Parri7696f992018-07-16 11:06:03 -07002209If a wakeup does occur, one (at least) of the two loads must see 1. If, on
2210the other hand, a wakeup does not occur, both loads might see 0.
2211
2212wake_up_process() always executes a general memory barrier. The barrier again
2213occurs before the task state is accessed. In particular, if the wake_up() in
2214the previous snippet were replaced by a call to wake_up_process() then one of
2215the two loads would be guaranteed to see 1.
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002216
David Howells50fa6102009-04-28 15:01:38 +01002217The available waker functions include:
2218
2219 complete();
2220 wake_up();
2221 wake_up_all();
2222 wake_up_bit();
2223 wake_up_interruptible();
2224 wake_up_interruptible_all();
2225 wake_up_interruptible_nr();
2226 wake_up_interruptible_poll();
2227 wake_up_interruptible_sync();
2228 wake_up_interruptible_sync_poll();
2229 wake_up_locked();
2230 wake_up_locked_poll();
2231 wake_up_nr();
2232 wake_up_poll();
2233 wake_up_process();
2234
Andrea Parri7696f992018-07-16 11:06:03 -07002235In terms of memory ordering, these functions all provide the same guarantees of
2236a wake_up() (or stronger).
David Howells50fa6102009-04-28 15:01:38 +01002237
2238[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2239order multiple stores before the wake-up with respect to loads of those stored
2240values after the sleeper has called set_current_state(). For instance, if the
2241sleeper does:
2242
2243 set_current_state(TASK_INTERRUPTIBLE);
2244 if (event_indicated)
2245 break;
2246 __set_current_state(TASK_RUNNING);
2247 do_something(my_data);
2248
2249and the waker does:
2250
2251 my_data = value;
2252 event_indicated = 1;
2253 wake_up(&event_wait_queue);
2254
2255there's no guarantee that the change to event_indicated will be perceived by
2256the sleeper as coming after the change to my_data. In such a circumstance, the
2257code on both sides must interpolate its own memory barriers between the
2258separate data accesses. Thus the above sleeper ought to do:
2259
2260 set_current_state(TASK_INTERRUPTIBLE);
2261 if (event_indicated) {
2262 smp_rmb();
2263 do_something(my_data);
2264 }
2265
2266and the waker should do:
2267
2268 my_data = value;
2269 smp_wmb();
2270 event_indicated = 1;
2271 wake_up(&event_wait_queue);
2272
2273
David Howells108b42b2006-03-31 16:00:29 +01002274MISCELLANEOUS FUNCTIONS
2275-----------------------
2276
2277Other functions that imply barriers:
2278
2279 (*) schedule() and similar imply full memory barriers.
2280
David Howells108b42b2006-03-31 16:00:29 +01002281
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002282===================================
2283INTER-CPU ACQUIRING BARRIER EFFECTS
2284===================================
David Howells108b42b2006-03-31 16:00:29 +01002285
2286On SMP systems locking primitives give a more substantial form of barrier: one
2287that does affect memory access ordering on other CPUs, within the context of
2288conflict on any particular lock.
2289
2290
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002291ACQUIRES VS MEMORY ACCESSES
2292---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002293
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002294Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002295three CPUs; then should the following sequence of events occur:
2296
2297 CPU 1 CPU 2
2298 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002299 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002300 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002301 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2302 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002303 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002304 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002305
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002306Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002307through *H occur in, other than the constraints imposed by the separate locks
SeongJae Park0b6fa342016-04-12 08:52:53 -07002308on the separate CPUs. It might, for example, see:
David Howells108b42b2006-03-31 16:00:29 +01002309
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002310 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002311
2312But it won't see any of:
2313
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002314 *B, *C or *D preceding ACQUIRE M
2315 *A, *B or *C following RELEASE M
2316 *F, *G or *H preceding ACQUIRE Q
2317 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002318
2319
David Howells108b42b2006-03-31 16:00:29 +01002320
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002321ACQUIRES VS I/O ACCESSES
2322------------------------
David Howells108b42b2006-03-31 16:00:29 +01002323
2324Under certain circumstances (especially involving NUMA), I/O accesses within
2325two spinlocked sections on two different CPUs may be seen as interleaved by the
2326PCI bridge, because the PCI bridge does not necessarily participate in the
2327cache-coherence protocol, and is therefore incapable of issuing the required
2328read memory barriers.
2329
2330For example:
2331
2332 CPU 1 CPU 2
2333 =============================== ===============================
2334 spin_lock(Q)
2335 writel(0, ADDR)
2336 writel(1, DATA);
2337 spin_unlock(Q);
2338 spin_lock(Q);
2339 writel(4, ADDR);
2340 writel(5, DATA);
2341 spin_unlock(Q);
2342
2343may be seen by the PCI bridge as follows:
2344
2345 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2346
2347which would probably cause the hardware to malfunction.
2348
2349
2350What is necessary here is to intervene with an mmiowb() before dropping the
2351spinlock, for example:
2352
2353 CPU 1 CPU 2
2354 =============================== ===============================
2355 spin_lock(Q)
2356 writel(0, ADDR)
2357 writel(1, DATA);
2358 mmiowb();
2359 spin_unlock(Q);
2360 spin_lock(Q);
2361 writel(4, ADDR);
2362 writel(5, DATA);
2363 mmiowb();
2364 spin_unlock(Q);
2365
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002366this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2367before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002368
2369
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002370Furthermore, following a store by a load from the same device obviates the need
2371for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002372is performed:
2373
2374 CPU 1 CPU 2
2375 =============================== ===============================
2376 spin_lock(Q)
2377 writel(0, ADDR)
2378 a = readl(DATA);
2379 spin_unlock(Q);
2380 spin_lock(Q);
2381 writel(4, ADDR);
2382 b = readl(DATA);
2383 spin_unlock(Q);
2384
2385
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002386See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002387
2388
2389=================================
2390WHERE ARE MEMORY BARRIERS NEEDED?
2391=================================
2392
2393Under normal operation, memory operation reordering is generally not going to
2394be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002395work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002396circumstances in which reordering definitely _could_ be a problem:
2397
2398 (*) Interprocessor interaction.
2399
2400 (*) Atomic operations.
2401
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002402 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002403
2404 (*) Interrupts.
2405
2406
2407INTERPROCESSOR INTERACTION
2408--------------------------
2409
2410When there's a system with more than one processor, more than one CPU in the
2411system may be working on the same data set at the same time. This can cause
2412synchronisation problems, and the usual way of dealing with them is to use
2413locks. Locks, however, are quite expensive, and so it may be preferable to
2414operate without the use of a lock if at all possible. In such a case
2415operations that affect both CPUs may have to be carefully ordered to prevent
2416a malfunction.
2417
2418Consider, for example, the R/W semaphore slow path. Here a waiting process is
2419queued on the semaphore, by virtue of it having a piece of its stack linked to
2420the semaphore's list of waiting processes:
2421
2422 struct rw_semaphore {
2423 ...
2424 spinlock_t lock;
2425 struct list_head waiters;
2426 };
2427
2428 struct rwsem_waiter {
2429 struct list_head list;
2430 struct task_struct *task;
2431 };
2432
2433To wake up a particular waiter, the up_read() or up_write() functions have to:
2434
2435 (1) read the next pointer from this waiter's record to know as to where the
2436 next waiter record is;
2437
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002438 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002439
2440 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2441
2442 (4) call wake_up_process() on the task; and
2443
2444 (5) release the reference held on the waiter's task struct.
2445
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002446In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002447
2448 LOAD waiter->list.next;
2449 LOAD waiter->task;
2450 STORE waiter->task;
2451 CALL wakeup
2452 RELEASE task
2453
2454and if any of these steps occur out of order, then the whole thing may
2455malfunction.
2456
2457Once it has queued itself and dropped the semaphore lock, the waiter does not
2458get the lock again; it instead just waits for its task pointer to be cleared
2459before proceeding. Since the record is on the waiter's stack, this means that
2460if the task pointer is cleared _before_ the next pointer in the list is read,
2461another CPU might start processing the waiter and might clobber the waiter's
2462stack before the up*() function has a chance to read the next pointer.
2463
2464Consider then what might happen to the above sequence of events:
2465
2466 CPU 1 CPU 2
2467 =============================== ===============================
2468 down_xxx()
2469 Queue waiter
2470 Sleep
2471 up_yyy()
2472 LOAD waiter->task;
2473 STORE waiter->task;
2474 Woken up by other event
2475 <preempt>
2476 Resume processing
2477 down_xxx() returns
2478 call foo()
2479 foo() clobbers *waiter
2480 </preempt>
2481 LOAD waiter->list.next;
2482 --- OOPS ---
2483
2484This could be dealt with using the semaphore lock, but then the down_xxx()
2485function has to needlessly get the spinlock again after being woken up.
2486
2487The way to deal with this is to insert a general SMP memory barrier:
2488
2489 LOAD waiter->list.next;
2490 LOAD waiter->task;
2491 smp_mb();
2492 STORE waiter->task;
2493 CALL wakeup
2494 RELEASE task
2495
2496In this case, the barrier makes a guarantee that all memory accesses before the
2497barrier will appear to happen before all the memory accesses after the barrier
2498with respect to the other CPUs on the system. It does _not_ guarantee that all
2499the memory accesses before the barrier will be complete by the time the barrier
2500instruction itself is complete.
2501
2502On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2503compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002504right order without actually intervening in the CPU. Since there's only one
2505CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002506
2507
2508ATOMIC OPERATIONS
2509-----------------
2510
Will Deacon806654a2018-11-19 11:02:45 +00002511While they are technically interprocessor interaction considerations, atomic
David Howellsdbc87002006-04-10 22:54:23 -07002512operations are noted specially as some of them imply full memory barriers and
2513some don't, but they're very heavily relied on as a group throughout the
2514kernel.
2515
Peter Zijlstra706eeb32017-06-12 14:50:27 +02002516See Documentation/atomic_t.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01002517
2518
2519ACCESSING DEVICES
2520-----------------
2521
2522Many devices can be memory mapped, and so appear to the CPU as if they're just
2523a set of memory locations. To control such a device, the driver usually has to
2524make the right memory accesses in exactly the right order.
2525
2526However, having a clever CPU or a clever compiler creates a potential problem
2527in that the carefully sequenced accesses in the driver code won't reach the
2528device in the requisite order if the CPU or the compiler thinks it is more
2529efficient to reorder, combine or merge accesses - something that would cause
2530the device to malfunction.
2531
2532Inside of the Linux kernel, I/O should be done through the appropriate accessor
2533routines - such as inb() or writel() - which know how to make such accesses
Will Deacon806654a2018-11-19 11:02:45 +00002534appropriately sequential. While this, for the most part, renders the explicit
David Howells108b42b2006-03-31 16:00:29 +01002535use of memory barriers unnecessary, there are a couple of situations where they
2536might be needed:
2537
2538 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2539 so for _all_ general drivers locks should be used and mmiowb() must be
2540 issued prior to unlocking the critical section.
2541
2542 (2) If the accessor functions are used to refer to an I/O memory window with
2543 relaxed memory access properties, then _mandatory_ memory barriers are
2544 required to enforce ordering.
2545
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002546See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002547
2548
2549INTERRUPTS
2550----------
2551
2552A driver may be interrupted by its own interrupt service routine, and thus the
2553two parts of the driver may interfere with each other's attempts to control or
2554access the device.
2555
2556This may be alleviated - at least in part - by disabling local interrupts (a
2557form of locking), such that the critical operations are all contained within
Will Deacon806654a2018-11-19 11:02:45 +00002558the interrupt-disabled section in the driver. While the driver's interrupt
David Howells108b42b2006-03-31 16:00:29 +01002559routine is executing, the driver's core may not run on the same CPU, and its
2560interrupt is not permitted to happen again until the current interrupt has been
2561handled, thus the interrupt handler does not need to lock against that.
2562
2563However, consider a driver that was talking to an ethernet card that sports an
2564address register and a data register. If that driver's core talks to the card
2565under interrupt-disablement and then the driver's interrupt handler is invoked:
2566
2567 LOCAL IRQ DISABLE
2568 writew(ADDR, 3);
2569 writew(DATA, y);
2570 LOCAL IRQ ENABLE
2571 <interrupt>
2572 writew(ADDR, 4);
2573 q = readw(DATA);
2574 </interrupt>
2575
2576The store to the data register might happen after the second store to the
2577address register if ordering rules are sufficiently relaxed:
2578
2579 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2580
2581
2582If ordering rules are relaxed, it must be assumed that accesses done inside an
2583interrupt disabled section may leak outside of it and may interleave with
2584accesses performed in an interrupt - and vice versa - unless implicit or
2585explicit barriers are used.
2586
2587Normally this won't be a problem because the I/O accesses done inside such
2588sections will include synchronous load operations on strictly ordered I/O
SeongJae Park0b6fa342016-04-12 08:52:53 -07002589registers that form implicit I/O barriers. If this isn't sufficient then an
David Howells108b42b2006-03-31 16:00:29 +01002590mmiowb() may need to be used explicitly.
2591
2592
2593A similar situation may occur between an interrupt routine and two routines
SeongJae Park0b6fa342016-04-12 08:52:53 -07002594running on separate CPUs that communicate with each other. If such a case is
David Howells108b42b2006-03-31 16:00:29 +01002595likely, then interrupt-disabling locks should be used to guarantee ordering.
2596
2597
2598==========================
2599KERNEL I/O BARRIER EFFECTS
2600==========================
2601
2602When accessing I/O memory, drivers should use the appropriate accessor
2603functions:
2604
2605 (*) inX(), outX():
2606
2607 These are intended to talk to I/O space rather than memory space, but
SeongJae Park0b6fa342016-04-12 08:52:53 -07002608 that's primarily a CPU-specific concept. The i386 and x86_64 processors
2609 do indeed have special I/O space access cycles and instructions, but many
David Howells108b42b2006-03-31 16:00:29 +01002610 CPUs don't have such a concept.
2611
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002612 The PCI bus, amongst others, defines an I/O space concept which - on such
2613 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002614 space. However, it may also be mapped as a virtual I/O space in the CPU's
2615 memory map, particularly on those CPUs that don't support alternate I/O
2616 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002617
2618 Accesses to this space may be fully synchronous (as on i386), but
2619 intermediary bridges (such as the PCI host bridge) may not fully honour
2620 that.
2621
2622 They are guaranteed to be fully ordered with respect to each other.
2623
2624 They are not guaranteed to be fully ordered with respect to other types of
2625 memory and I/O operation.
2626
2627 (*) readX(), writeX():
2628
2629 Whether these are guaranteed to be fully ordered and uncombined with
2630 respect to each other on the issuing CPU depends on the characteristics
SeongJae Park0b6fa342016-04-12 08:52:53 -07002631 defined for the memory window through which they're accessing. On later
David Howells108b42b2006-03-31 16:00:29 +01002632 i386 architecture machines, for example, this is controlled by way of the
2633 MTRR registers.
2634
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002635 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002636 provided they're not accessing a prefetchable device.
2637
2638 However, intermediary hardware (such as a PCI bridge) may indulge in
2639 deferral if it so wishes; to flush a store, a load from the same location
2640 is preferred[*], but a load from the same device or from configuration
2641 space should suffice for PCI.
2642
2643 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002644 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2645 example.
David Howells108b42b2006-03-31 16:00:29 +01002646
2647 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2648 force stores to be ordered.
2649
2650 Please refer to the PCI specification for more information on interactions
2651 between PCI transactions.
2652
Will Deacona8e0aea2013-09-04 12:30:08 +01002653 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002654
Will Deacona8e0aea2013-09-04 12:30:08 +01002655 These are similar to readX() and writeX(), but provide weaker memory
SeongJae Park0b6fa342016-04-12 08:52:53 -07002656 ordering guarantees. Specifically, they do not guarantee ordering with
Will Deacona8e0aea2013-09-04 12:30:08 +01002657 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
SeongJae Park0b6fa342016-04-12 08:52:53 -07002658 ordering with respect to LOCK or UNLOCK operations. If the latter is
2659 required, an mmiowb() barrier can be used. Note that relaxed accesses to
Will Deacona8e0aea2013-09-04 12:30:08 +01002660 the same peripheral are guaranteed to be ordered with respect to each
2661 other.
David Howells108b42b2006-03-31 16:00:29 +01002662
2663 (*) ioreadX(), iowriteX()
2664
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002665 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002666 doing, be it inX()/outX() or readX()/writeX().
2667
2668
2669========================================
2670ASSUMED MINIMUM EXECUTION ORDERING MODEL
2671========================================
2672
2673It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2674maintain the appearance of program causality with respect to itself. Some CPUs
2675(such as i386 or x86_64) are more constrained than others (such as powerpc or
2676frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2677of arch-specific code.
2678
2679This means that it must be considered that the CPU will execute its instruction
2680stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002681instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002682earlier instruction must be sufficiently complete[*] before the later
2683instruction may proceed; in other words: provided that the appearance of
2684causality is maintained.
2685
2686 [*] Some instructions have more than one effect - such as changing the
2687 condition codes, changing registers or changing memory - and different
2688 instructions may depend on different effects.
2689
2690A CPU may also discard any instruction sequence that winds up having no
2691ultimate effect. For example, if two adjacent instructions both load an
2692immediate value into the same register, the first may be discarded.
2693
2694
2695Similarly, it has to be assumed that compiler might reorder the instruction
2696stream in any way it sees fit, again provided the appearance of causality is
2697maintained.
2698
2699
2700============================
2701THE EFFECTS OF THE CPU CACHE
2702============================
2703
2704The way cached memory operations are perceived across the system is affected to
2705a certain extent by the caches that lie between CPUs and memory, and by the
2706memory coherence system that maintains the consistency of state in the system.
2707
2708As far as the way a CPU interacts with another part of the system through the
2709caches goes, the memory system has to include the CPU's caches, and memory
2710barriers for the most part act at the interface between the CPU and its cache
2711(memory barriers logically act on the dotted line in the following diagram):
2712
2713 <--- CPU ---> : <----------- Memory ----------->
2714 :
2715 +--------+ +--------+ : +--------+ +-----------+
2716 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002717 | CPU | | Memory | : | CPU | | | | |
2718 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002719 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002720 | | | | : | | | | | |
2721 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002722 : | Cache | +--------+
2723 : | Coherency |
2724 : | Mechanism | +--------+
2725 +--------+ +--------+ : +--------+ | | | |
2726 | | | | : | | | | | |
2727 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002728 | Core |--->| Access |----->| Cache |<-->| | | |
2729 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002730 | | | | : | | | | +--------+
2731 +--------+ +--------+ : +--------+ +-----------+
2732 :
2733 :
2734
2735Although any particular load or store may not actually appear outside of the
2736CPU that issued it since it may have been satisfied within the CPU's own cache,
2737it will still appear as if the full memory access had taken place as far as the
2738other CPUs are concerned since the cache coherency mechanisms will migrate the
2739cacheline over to the accessing CPU and propagate the effects upon conflict.
2740
2741The CPU core may execute instructions in any order it deems fit, provided the
2742expected program causality appears to be maintained. Some of the instructions
2743generate load and store operations which then go into the queue of memory
2744accesses to be performed. The core may place these in the queue in any order
2745it wishes, and continue execution until it is forced to wait for an instruction
2746to complete.
2747
2748What memory barriers are concerned with is controlling the order in which
2749accesses cross from the CPU side of things to the memory side of things, and
2750the order in which the effects are perceived to happen by the other observers
2751in the system.
2752
2753[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2754their own loads and stores as if they had happened in program order.
2755
2756[!] MMIO or other device accesses may bypass the cache system. This depends on
2757the properties of the memory window through which devices are accessed and/or
2758the use of any special device communication instructions the CPU may have.
2759
2760
2761CACHE COHERENCY
2762---------------
2763
2764Life isn't quite as simple as it may appear above, however: for while the
2765caches are expected to be coherent, there's no guarantee that that coherency
Will Deacon806654a2018-11-19 11:02:45 +00002766will be ordered. This means that while changes made on one CPU will
David Howells108b42b2006-03-31 16:00:29 +01002767eventually become visible on all CPUs, there's no guarantee that they will
2768become apparent in the same order on those other CPUs.
2769
2770
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002771Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2772has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002773
2774 :
2775 : +--------+
2776 : +---------+ | |
2777 +--------+ : +--->| Cache A |<------->| |
2778 | | : | +---------+ | |
2779 | CPU 1 |<---+ | |
2780 | | : | +---------+ | |
2781 +--------+ : +--->| Cache B |<------->| |
2782 : +---------+ | |
2783 : | Memory |
2784 : +---------+ | System |
2785 +--------+ : +--->| Cache C |<------->| |
2786 | | : | +---------+ | |
2787 | CPU 2 |<---+ | |
2788 | | : | +---------+ | |
2789 +--------+ : +--->| Cache D |<------->| |
2790 : +---------+ | |
2791 : +--------+
2792 :
2793
2794Imagine the system has the following properties:
2795
2796 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2797 resident in memory;
2798
2799 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2800 resident in memory;
2801
Will Deacon806654a2018-11-19 11:02:45 +00002802 (*) while the CPU core is interrogating one cache, the other cache may be
David Howells108b42b2006-03-31 16:00:29 +01002803 making use of the bus to access the rest of the system - perhaps to
2804 displace a dirty cacheline or to do a speculative load;
2805
2806 (*) each cache has a queue of operations that need to be applied to that cache
2807 to maintain coherency with the rest of the system;
2808
2809 (*) the coherency queue is not flushed by normal loads to lines already
2810 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002811 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002812
2813Imagine, then, that two writes are made on the first CPU, with a write barrier
2814between them to guarantee that they will appear to reach that CPU's caches in
2815the requisite order:
2816
2817 CPU 1 CPU 2 COMMENT
2818 =============== =============== =======================================
2819 u == 0, v == 1 and p == &u, q == &u
2820 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002821 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002822 change to p
2823 <A:modify v=2> v is now in cache A exclusively
2824 p = &v;
2825 <B:modify p=&v> p is now in cache B exclusively
2826
2827The write memory barrier forces the other CPUs in the system to perceive that
2828the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002829now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002830
2831 CPU 1 CPU 2 COMMENT
2832 =============== =============== =======================================
2833 ...
2834 q = p;
2835 x = *q;
2836
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002837The above pair of reads may then fail to happen in the expected order, as the
Will Deacon806654a2018-11-19 11:02:45 +00002838cacheline holding p may get updated in one of the second CPU's caches while
David Howells108b42b2006-03-31 16:00:29 +01002839the update to the cacheline holding v is delayed in the other of the second
2840CPU's caches by some other cache event:
2841
2842 CPU 1 CPU 2 COMMENT
2843 =============== =============== =======================================
2844 u == 0, v == 1 and p == &u, q == &u
2845 v = 2;
2846 smp_wmb();
2847 <A:modify v=2> <C:busy>
2848 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002849 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002850 <D:request p>
2851 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002852 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002853 x = *q;
2854 <C:read *q> Reads from v before v updated in cache
2855 <C:unbusy>
2856 <C:commit v=2>
2857
Will Deacon806654a2018-11-19 11:02:45 +00002858Basically, while both cachelines will be updated on CPU 2 eventually, there's
David Howells108b42b2006-03-31 16:00:29 +01002859no guarantee that, without intervention, the order of update will be the same
2860as that committed on CPU 1.
2861
2862
2863To intervene, we need to interpolate a data dependency barrier or a read
Paul E. McKenneyf28f0862018-03-07 09:27:37 -08002864barrier between the loads (which as of v4.15 is supplied unconditionally
2865by the READ_ONCE() macro). This will force the cache to commit its
2866coherency queue before processing any further requests:
David Howells108b42b2006-03-31 16:00:29 +01002867
2868 CPU 1 CPU 2 COMMENT
2869 =============== =============== =======================================
2870 u == 0, v == 1 and p == &u, q == &u
2871 v = 2;
2872 smp_wmb();
2873 <A:modify v=2> <C:busy>
2874 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002875 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002876 <D:request p>
2877 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002878 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002879 smp_read_barrier_depends()
2880 <C:unbusy>
2881 <C:commit v=2>
2882 x = *q;
2883 <C:read *q> Reads from v after v updated in cache
2884
2885
2886This sort of problem can be encountered on DEC Alpha processors as they have a
2887split cache that improves performance by making better use of the data bus.
Will Deacon806654a2018-11-19 11:02:45 +00002888While most CPUs do imply a data dependency barrier on the read when a memory
David Howells108b42b2006-03-31 16:00:29 +01002889access depends on a read, not all do, so it may not be relied on.
2890
2891Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002892cachelets for normal memory accesses. The semantics of the Alpha removes the
Paul E. McKenney9ad3c142017-11-27 09:20:40 -08002893need for hardware coordination in the absence of memory barriers, which
2894permitted Alpha to sport higher CPU clock rates back in the day. However,
Paul E. McKenneyf28f0862018-03-07 09:27:37 -08002895please note that (again, as of v4.15) smp_read_barrier_depends() should not
2896be used except in Alpha arch-specific code and within the READ_ONCE() macro.
David Howells108b42b2006-03-31 16:00:29 +01002897
2898
2899CACHE COHERENCY VS DMA
2900----------------------
2901
2902Not all systems maintain cache coherency with respect to devices doing DMA. In
2903such cases, a device attempting DMA may obtain stale data from RAM because
2904dirty cache lines may be resident in the caches of various CPUs, and may not
2905have been written back to RAM yet. To deal with this, the appropriate part of
2906the kernel must flush the overlapping bits of cache on each CPU (and maybe
2907invalidate them as well).
2908
2909In addition, the data DMA'd to RAM by a device may be overwritten by dirty
2910cache lines being written back to RAM from a CPU's cache after the device has
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002911installed its own data, or cache lines present in the CPU's cache may simply
2912obscure the fact that RAM has been updated, until at such time as the cacheline
2913is discarded from the CPU's cache and reloaded. To deal with this, the
2914appropriate part of the kernel must invalidate the overlapping bits of the
David Howells108b42b2006-03-31 16:00:29 +01002915cache on each CPU.
2916
Mauro Carvalho Chehabde0f51e2018-05-07 06:35:41 -03002917See Documentation/core-api/cachetlb.rst for more information on cache management.
David Howells108b42b2006-03-31 16:00:29 +01002918
2919
2920CACHE COHERENCY VS MMIO
2921-----------------------
2922
2923Memory mapped I/O usually takes place through memory locations that are part of
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002924a window in the CPU's memory space that has different properties assigned than
David Howells108b42b2006-03-31 16:00:29 +01002925the usual RAM directed window.
2926
2927Amongst these properties is usually the fact that such accesses bypass the
2928caching entirely and go directly to the device buses. This means MMIO accesses
2929may, in effect, overtake accesses to cached memory that were emitted earlier.
2930A memory barrier isn't sufficient in such a case, but rather the cache must be
2931flushed between the cached memory write and the MMIO access if the two are in
2932any way dependent.
2933
2934
2935=========================
2936THE THINGS CPUS GET UP TO
2937=========================
2938
2939A programmer might take it for granted that the CPU will perform memory
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002940operations in exactly the order specified, so that if the CPU is, for example,
David Howells108b42b2006-03-31 16:00:29 +01002941given the following piece of code to execute:
2942
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002943 a = READ_ONCE(*A);
2944 WRITE_ONCE(*B, b);
2945 c = READ_ONCE(*C);
2946 d = READ_ONCE(*D);
2947 WRITE_ONCE(*E, e);
David Howells108b42b2006-03-31 16:00:29 +01002948
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002949they would then expect that the CPU will complete the memory operation for each
David Howells108b42b2006-03-31 16:00:29 +01002950instruction before moving on to the next one, leading to a definite sequence of
2951operations as seen by external observers in the system:
2952
2953 LOAD *A, STORE *B, LOAD *C, LOAD *D, STORE *E.
2954
2955
2956Reality is, of course, much messier. With many CPUs and compilers, the above
2957assumption doesn't hold because:
2958
2959 (*) loads are more likely to need to be completed immediately to permit
2960 execution progress, whereas stores can often be deferred without a
2961 problem;
2962
2963 (*) loads may be done speculatively, and the result discarded should it prove
2964 to have been unnecessary;
2965
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002966 (*) loads may be done speculatively, leading to the result having been fetched
2967 at the wrong time in the expected sequence of events;
David Howells108b42b2006-03-31 16:00:29 +01002968
2969 (*) the order of the memory accesses may be rearranged to promote better use
2970 of the CPU buses and caches;
2971
2972 (*) loads and stores may be combined to improve performance when talking to
2973 memory or I/O hardware that can do batched accesses of adjacent locations,
2974 thus cutting down on transaction setup costs (memory and PCI devices may
2975 both be able to do this); and
2976
Will Deacon806654a2018-11-19 11:02:45 +00002977 (*) the CPU's data cache may affect the ordering, and while cache-coherency
David Howells108b42b2006-03-31 16:00:29 +01002978 mechanisms may alleviate this - once the store has actually hit the cache
2979 - there's no guarantee that the coherency management will be propagated in
2980 order to other CPUs.
2981
2982So what another CPU, say, might actually observe from the above piece of code
2983is:
2984
2985 LOAD *A, ..., LOAD {*C,*D}, STORE *E, STORE *B
2986
2987 (Where "LOAD {*C,*D}" is a combined load)
2988
2989
2990However, it is guaranteed that a CPU will be self-consistent: it will see its
2991_own_ accesses appear to be correctly ordered, without the need for a memory
2992barrier. For instance with the following code:
2993
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002994 U = READ_ONCE(*A);
2995 WRITE_ONCE(*A, V);
2996 WRITE_ONCE(*A, W);
2997 X = READ_ONCE(*A);
2998 WRITE_ONCE(*A, Y);
2999 Z = READ_ONCE(*A);
David Howells108b42b2006-03-31 16:00:29 +01003000
3001and assuming no intervention by an external influence, it can be assumed that
3002the final result will appear to be:
3003
3004 U == the original value of *A
3005 X == W
3006 Z == Y
3007 *A == Y
3008
3009The code above may cause the CPU to generate the full sequence of memory
3010accesses:
3011
3012 U=LOAD *A, STORE *A=V, STORE *A=W, X=LOAD *A, STORE *A=Y, Z=LOAD *A
3013
3014in that order, but, without intervention, the sequence may have almost any
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003015combination of elements combined or discarded, provided the program's view
3016of the world remains consistent. Note that READ_ONCE() and WRITE_ONCE()
3017are -not- optional in the above example, as there are architectures
3018where a given CPU might reorder successive loads to the same location.
3019On such architectures, READ_ONCE() and WRITE_ONCE() do whatever is
3020necessary to prevent this, for example, on Itanium the volatile casts
3021used by READ_ONCE() and WRITE_ONCE() cause GCC to emit the special ld.acq
3022and st.rel instructions (respectively) that prevent such reordering.
David Howells108b42b2006-03-31 16:00:29 +01003023
3024The compiler may also combine, discard or defer elements of the sequence before
3025the CPU even sees them.
3026
3027For instance:
3028
3029 *A = V;
3030 *A = W;
3031
3032may be reduced to:
3033
3034 *A = W;
3035
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003036since, without either a write barrier or an WRITE_ONCE(), it can be
Paul E. McKenney2ecf8102013-12-11 13:59:04 -08003037assumed that the effect of the storage of V to *A is lost. Similarly:
David Howells108b42b2006-03-31 16:00:29 +01003038
3039 *A = Y;
3040 Z = *A;
3041
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003042may, without a memory barrier or an READ_ONCE() and WRITE_ONCE(), be
3043reduced to:
David Howells108b42b2006-03-31 16:00:29 +01003044
3045 *A = Y;
3046 Z = Y;
3047
3048and the LOAD operation never appear outside of the CPU.
3049
3050
3051AND THEN THERE'S THE ALPHA
3052--------------------------
3053
3054The DEC Alpha CPU is one of the most relaxed CPUs there is. Not only that,
3055some versions of the Alpha CPU have a split data cache, permitting them to have
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003056two semantically-related cache lines updated at separate times. This is where
David Howells108b42b2006-03-31 16:00:29 +01003057the data dependency barrier really becomes necessary as this synchronises both
3058caches with the memory coherence system, thus making it seem like pointer
3059changes vs new data occur in the right order.
3060
Paul E. McKenneyf28f0862018-03-07 09:27:37 -08003061The Alpha defines the Linux kernel's memory model, although as of v4.15
3062the Linux kernel's addition of smp_read_barrier_depends() to READ_ONCE()
3063greatly reduced Alpha's impact on the memory model.
David Howells108b42b2006-03-31 16:00:29 +01003064
3065See the subsection on "Cache Coherency" above.
3066
SeongJae Park0b6fa342016-04-12 08:52:53 -07003067
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003068VIRTUAL MACHINE GUESTS
SeongJae Park3dbf0912016-04-12 08:52:52 -07003069----------------------
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003070
3071Guests running within virtual machines might be affected by SMP effects even if
3072the guest itself is compiled without SMP support. This is an artifact of
3073interfacing with an SMP host while running an UP kernel. Using mandatory
3074barriers for this use-case would be possible but is often suboptimal.
3075
3076To handle this case optimally, low-level virt_mb() etc macros are available.
3077These have the same effect as smp_mb() etc when SMP is enabled, but generate
SeongJae Park0b6fa342016-04-12 08:52:53 -07003078identical code for SMP and non-SMP systems. For example, virtual machine guests
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003079should use virt_mb() rather than smp_mb() when synchronizing against a
3080(possibly SMP) host.
3081
3082These are equivalent to smp_mb() etc counterparts in all other respects,
3083in particular, they do not control MMIO effects: to control
3084MMIO effects, use mandatory barriers.
David Howells108b42b2006-03-31 16:00:29 +01003085
SeongJae Park0b6fa342016-04-12 08:52:53 -07003086
David Howells90fddab2010-03-24 09:43:00 +00003087============
3088EXAMPLE USES
3089============
3090
3091CIRCULAR BUFFERS
3092----------------
3093
3094Memory barriers can be used to implement circular buffering without the need
3095of a lock to serialise the producer with the consumer. See:
3096
Mauro Carvalho Chehabd8a121e2018-05-07 06:35:43 -03003097 Documentation/core-api/circular-buffers.rst
David Howells90fddab2010-03-24 09:43:00 +00003098
3099for details.
3100
3101
David Howells108b42b2006-03-31 16:00:29 +01003102==========
3103REFERENCES
3104==========
3105
3106Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
3107Digital Press)
3108 Chapter 5.2: Physical Address Space Characteristics
3109 Chapter 5.4: Caches and Write Buffers
3110 Chapter 5.5: Data Sharing
3111 Chapter 5.6: Read/Write Ordering
3112
3113AMD64 Architecture Programmer's Manual Volume 2: System Programming
3114 Chapter 7.1: Memory-Access Ordering
3115 Chapter 7.4: Buffering and Combining Memory Writes
3116
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07003117ARM Architecture Reference Manual (ARMv8, for ARMv8-A architecture profile)
3118 Chapter B2: The AArch64 Application Level Memory Model
3119
David Howells108b42b2006-03-31 16:00:29 +01003120IA-32 Intel Architecture Software Developer's Manual, Volume 3:
3121System Programming Guide
3122 Chapter 7.1: Locked Atomic Operations
3123 Chapter 7.2: Memory Ordering
3124 Chapter 7.4: Serializing Instructions
3125
3126The SPARC Architecture Manual, Version 9
3127 Chapter 8: Memory Models
3128 Appendix D: Formal Specification of the Memory Models
3129 Appendix J: Programming with the Memory Models
3130
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07003131Storage in the PowerPC (Stone and Fitzgerald)
3132
David Howells108b42b2006-03-31 16:00:29 +01003133UltraSPARC Programmer Reference Manual
3134 Chapter 5: Memory Accesses and Cacheability
3135 Chapter 15: Sparc-V9 Memory Models
3136
3137UltraSPARC III Cu User's Manual
3138 Chapter 9: Memory Models
3139
3140UltraSPARC IIIi Processor User's Manual
3141 Chapter 8: Memory Models
3142
3143UltraSPARC Architecture 2005
3144 Chapter 9: Memory
3145 Appendix D: Formal Specifications of the Memory Models
3146
3147UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3148 Chapter 8: Memory Models
3149 Appendix F: Caches and Cache Coherency
3150
3151Solaris Internals, Core Kernel Architecture, p63-68:
3152 Chapter 3.3: Hardware Considerations for Locks and
3153 Synchronization
3154
3155Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3156for Kernel Programmers:
3157 Chapter 13: Other Memory Models
3158
3159Intel Itanium Architecture Software Developer's Manual: Volume 1:
3160 Section 2.6: Speculation
3161 Section 4.4: Memory Access