blob: a37d3aff3e739bd89bdb83f98571c72dddefb8c5 [file] [log] [blame]
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?
55 - Data dependency barriers.
56 - 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
406 make sure that the target of the second load is updated before the address
407 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()
474 and smp_cond_acquire() operations. The later builds the necessary ACQUIRE
475 semantics from relying on a control dependency and smp_rmb().
David Howells108b42b2006-03-31 16:00:29 +0100476
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100477 Memory operations that occur before an ACQUIRE operation may appear to
478 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100479
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100480 An ACQUIRE operation should almost always be paired with a RELEASE
481 operation.
David Howells108b42b2006-03-31 16:00:29 +0100482
483
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100484 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100485
486 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100487 memory operations before the RELEASE operation will appear to happen
488 before the RELEASE operation with respect to the other components of the
489 system. RELEASE operations include UNLOCK operations and
490 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100491
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100492 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100493 happen before it completes.
494
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100495 The use of ACQUIRE and RELEASE operations generally precludes the need
496 for other sorts of memory barrier (but note the exceptions mentioned in
497 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
498 pair is -not- guaranteed to act as a full memory barrier. However, after
499 an ACQUIRE on a given variable, all memory accesses preceding any prior
500 RELEASE on that same variable are guaranteed to be visible. In other
501 words, within a given variable's critical section, all accesses of all
502 previous critical sections for that variable are guaranteed to have
503 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800504
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100505 This means that ACQUIRE acts as a minimal "acquire" operation and
506 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100507
Peter Zijlstra706eeb32017-06-12 14:50:27 +0200508A subset of the atomic operations described in atomic_t.txt have ACQUIRE and
509RELEASE variants in addition to fully-ordered and relaxed (no barrier
510semantics) definitions. For compound atomics performing both a load and a
511store, ACQUIRE semantics apply only to the load and RELEASE semantics apply
512only to the store portion of the operation.
David Howells108b42b2006-03-31 16:00:29 +0100513
514Memory barriers are only required where there's a possibility of interaction
515between two CPUs or between a CPU and a device. If it can be guaranteed that
516there won't be any such interaction in any particular piece of code, then
517memory barriers are unnecessary in that piece of code.
518
519
520Note that these are the _minimum_ guarantees. Different architectures may give
521more substantial guarantees, but they may _not_ be relied upon outside of arch
522specific code.
523
524
525WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
526----------------------------------------------
527
528There are certain things that the Linux kernel memory barriers do not guarantee:
529
530 (*) There is no guarantee that any of the memory accesses specified before a
531 memory barrier will be _complete_ by the completion of a memory barrier
532 instruction; the barrier can be considered to draw a line in that CPU's
533 access queue that accesses of the appropriate type may not cross.
534
535 (*) There is no guarantee that issuing a memory barrier on one CPU will have
536 any direct effect on another CPU or any other hardware in the system. The
537 indirect effect will be the order in which the second CPU sees the effects
538 of the first CPU's accesses occur, but see the next point:
539
David Howells6bc39272006-06-25 05:49:22 -0700540 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100541 from a second CPU's accesses, even _if_ the second CPU uses a memory
542 barrier, unless the first CPU _also_ uses a matching memory barrier (see
543 the subsection on "SMP Barrier Pairing").
544
545 (*) There is no guarantee that some intervening piece of off-the-CPU
546 hardware[*] will not reorder the memory accesses. CPU cache coherency
547 mechanisms should propagate the indirect effects of a memory barrier
548 between CPUs, but might not do so in order.
549
550 [*] For information on bus mastering DMA and coherency please read:
551
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700552 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200553 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100554 Documentation/DMA-API.txt
555
556
557DATA DEPENDENCY BARRIERS
558------------------------
559
560The usage requirements of data dependency barriers are a little subtle, and
561it's not always obvious that they're needed. To illustrate, consider the
562following sequence of events:
563
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800564 CPU 1 CPU 2
565 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700566 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100567 B = 4;
568 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700569 WRITE_ONCE(P, &B)
570 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800571 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100572
573There's a clear data dependency here, and it would seem that by the end of the
574sequence, Q must be either &A or &B, and that:
575
576 (Q == &A) implies (D == 1)
577 (Q == &B) implies (D == 4)
578
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700579But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100580leading to the following situation:
581
582 (Q == &B) and (D == 2) ????
583
584Whilst this may seem like a failure of coherency or causality maintenance, it
585isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
586Alpha).
587
David Howells2b948952006-06-25 05:48:49 -0700588To deal with this, a data dependency barrier or better must be inserted
589between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100590
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800591 CPU 1 CPU 2
592 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700593 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100594 B = 4;
595 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700596 WRITE_ONCE(P, &B);
597 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800598 <data dependency barrier>
599 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100600
601This enforces the occurrence of one of the two implications, and prevents the
602third possibility from arising.
603
Paul E. McKenney92a84dd2016-01-14 14:17:04 -0800604
David Howells108b42b2006-03-31 16:00:29 +0100605[!] Note that this extremely counterintuitive situation arises most easily on
606machines with split caches, so that, for example, one cache bank processes
607even-numbered cache lines and the other bank processes odd-numbered cache
608lines. The pointer P might be stored in an odd-numbered cache line, and the
609variable B might be stored in an even-numbered cache line. Then, if the
610even-numbered bank of the reading CPU's cache is extremely busy while the
611odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700612but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100613
614
Paul E. McKenney66ce3a42017-06-30 16:18:28 -0700615A data-dependency barrier is not required to order dependent writes
616because the CPUs that the Linux kernel supports don't do writes
617until they are certain (1) that the write will actually happen, (2)
618of the location of the write, and (3) of the value to be written.
619But please carefully read the "CONTROL DEPENDENCIES" section and the
620Documentation/RCU/rcu_dereference.txt file: The compiler can and does
621break dependencies in a great many highly creative ways.
622
623 CPU 1 CPU 2
624 =============== ===============
625 { A == 1, B == 2, C = 3, P == &A, Q == &C }
626 B = 4;
627 <write barrier>
628 WRITE_ONCE(P, &B);
629 Q = READ_ONCE(P);
630 WRITE_ONCE(*Q, 5);
631
632Therefore, no data-dependency barrier is required to order the read into
633Q with the store into *Q. In other words, this outcome is prohibited,
634even without a data-dependency barrier:
635
636 (Q == &B) && (B == 4)
637
638Please note that this pattern should be rare. After all, the whole point
639of dependency ordering is to -prevent- writes to the data structure, along
640with the expensive cache misses associated with those writes. This pattern
641can be used to record rare error conditions and the like, and the CPUs'
642naturally occurring ordering prevents such records from being lost.
643
644
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700645Note well that the ordering provided by a data dependency is local to
646the CPU containing it. See the section on "Multicopy atomicity" for
647more information.
648
649
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800650The data dependency barrier is very important to the RCU system,
651for example. See rcu_assign_pointer() and rcu_dereference() in
652include/linux/rcupdate.h. This permits the current target of an RCU'd
653pointer to be replaced with a new modified target, without the replacement
654target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100655
656See also the subsection on "Cache Coherency" for a more thorough example.
657
658
659CONTROL DEPENDENCIES
660--------------------
661
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800662Control dependencies can be a bit tricky because current compilers do
663not understand them. The purpose of this section is to help you prevent
664the compiler's ignorance from breaking your code.
665
Paul E. McKenneyff382812015-02-17 10:00:06 -0800666A load-load control dependency requires a full read memory barrier, not
667simply a data dependency barrier to make it work correctly. Consider the
668following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100669
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700670 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800671 if (q) {
672 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700673 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700674 }
David Howells108b42b2006-03-31 16:00:29 +0100675
676This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800677dependency, but rather a control dependency that the CPU may short-circuit
678by attempting to predict the outcome in advance, so that other CPUs see
679the load from b as having happened before the load from a. In such a
680case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100681
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700682 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800683 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700684 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700685 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700686 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800687
688However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800689for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800690
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800691 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700692 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800693 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800694 }
695
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800696Control dependencies pair normally with other types of barriers.
697That said, please note that neither READ_ONCE() nor WRITE_ONCE()
698are optional! Without the READ_ONCE(), the compiler might combine the
699load from 'a' with other loads from 'a'. Without the WRITE_ONCE(),
700the compiler might combine the store to 'b' with other stores to 'b'.
701Either can result in highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800702
703Worse yet, if the compiler is able to prove (say) that the value of
704variable 'a' is always non-zero, it would be well within its rights
705to optimize the original example by eliminating the "if" statement
706as follows:
707
708 q = a;
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800709 b = 1; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800710
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800711So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700712
713It is tempting to try to enforce ordering on identical stores on both
714branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800715
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800716 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800717 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800718 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800719 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800720 do_something();
721 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800722 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800723 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800724 do_something_else();
725 }
726
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700727Unfortunately, current compilers will transform this as follows at high
728optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800729
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800730 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700731 barrier();
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800732 WRITE_ONCE(b, 1); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800733 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800734 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800735 do_something();
736 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800737 /* WRITE_ONCE(b, 1); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800738 do_something_else();
739 }
740
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700741Now there is no conditional between the load from 'a' and the store to
742'b', which means that the CPU is within its rights to reorder them:
743The conditional is absolutely required, and must be present in the
744assembly code even after all compiler optimizations have been applied.
745Therefore, if you need ordering in this example, you need explicit
746memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800747
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700748 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700749 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800750 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800751 do_something();
752 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800753 smp_store_release(&b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800754 do_something_else();
755 }
756
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700757In contrast, without explicit memory barriers, two-legged-if control
758ordering is guaranteed only when the stores differ, for example:
759
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800760 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700761 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800762 WRITE_ONCE(b, 1);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700763 do_something();
764 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800765 WRITE_ONCE(b, 2);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700766 do_something_else();
767 }
768
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800769The initial READ_ONCE() is still required to prevent the compiler from
770proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800771
772In addition, you need to be careful what you do with the local variable 'q',
773otherwise the compiler might be able to guess the value and again remove
774the needed conditional. For example:
775
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800776 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800777 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800778 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800779 do_something();
780 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800781 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800782 do_something_else();
783 }
784
785If MAX is defined to be 1, then the compiler knows that (q % MAX) is
786equal to zero, in which case the compiler is within its rights to
787transform the above code into the following:
788
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800789 q = READ_ONCE(a);
pierre Kuob26cfc42017-04-07 14:37:36 +0800790 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800791 do_something_else();
792
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700793Given this transformation, the CPU is not required to respect the ordering
794between the load from variable 'a' and the store to variable 'b'. It is
795tempting to add a barrier(), but this does not help. The conditional
796is gone, and the barrier won't bring it back. Therefore, if you are
797relying on this ordering, you should make sure that MAX is greater than
798one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800799
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800800 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800801 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
802 if (q % MAX) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800803 WRITE_ONCE(b, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800804 do_something();
805 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800806 WRITE_ONCE(b, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800807 do_something_else();
808 }
809
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700810Please note once again that the stores to 'b' differ. If they were
811identical, as noted earlier, the compiler could pull this store outside
812of the 'if' statement.
813
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700814You must also be careful not to rely too much on boolean short-circuit
815evaluation. Consider this example:
816
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800817 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700818 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700819 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700820
Paul E. McKenney5af46922015-04-25 12:48:29 -0700821Because the first condition cannot fault and the second condition is
822always true, the compiler can transform this example as following,
823defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700824
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800825 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700826 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700827
828This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700829out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700830the compiler to actually emit code for a given load, it does not force
831the compiler to use the results.
832
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700833In addition, control dependencies apply only to the then-clause and
834else-clause of the if-statement in question. In particular, it does
835not necessarily apply to code following the if-statement:
836
837 q = READ_ONCE(a);
838 if (q) {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800839 WRITE_ONCE(b, 1);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700840 } else {
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800841 WRITE_ONCE(b, 2);
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700842 }
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800843 WRITE_ONCE(c, 1); /* BUG: No ordering against the read from 'a'. */
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700844
845It is tempting to argue that there in fact is ordering because the
846compiler cannot reorder volatile accesses and also cannot reorder
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800847the writes to 'b' with the condition. Unfortunately for this line
848of reasoning, the compiler might compile the two writes to 'b' as
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700849conditional-move instructions, as in this fanciful pseudo-assembly
850language:
851
852 ld r1,a
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700853 cmp r1,$0
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800854 cmov,ne r4,$1
855 cmov,eq r4,$2
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700856 st r4,b
857 st $1,c
858
859A weakly ordered CPU would have no dependency of any sort between the load
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800860from 'a' and the store to 'c'. The control dependencies would extend
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700861only to the pair of cmov instructions and the store depending on them.
862In short, control dependencies apply only to the stores in the then-clause
863and else-clause of the if-statement in question (including functions
864invoked by those two clauses), not to code following that if-statement.
865
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800866
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700867Note well that the ordering provided by a control dependency is local
868to the CPU containing it. See the section on "Multicopy atomicity"
869for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800870
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800871
872In summary:
873
874 (*) Control dependencies can order prior loads against later stores.
875 However, they do -not- guarantee any other sort of ordering:
876 Not prior loads against later loads, nor prior stores against
877 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800878 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800879 later loads, smp_mb().
880
Paul E. McKenney7817b792015-12-29 16:23:18 -0800881 (*) If both legs of the "if" statement begin with identical stores to
882 the same variable, then those stores must be ordered, either by
883 preceding both of them with smp_mb() or by using smp_store_release()
884 to carry out the stores. Please note that it is -not- sufficient
Paul E. McKenneya5052652016-04-12 08:52:49 -0700885 to use barrier() at beginning of each leg of the "if" statement
886 because, as shown by the example above, optimizing compilers can
887 destroy the control dependency while respecting the letter of the
888 barrier() law.
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800889
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800890 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800891 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700892 conditional must involve the prior load. If the compiler is able
893 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800894 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
895 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800896
897 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800898 dependency into nonexistence. Careful use of READ_ONCE() or
899 atomic{,64}_read() can help to preserve your control dependency.
Paul E. McKenney895f5542016-01-06 14:23:03 -0800900 Please see the COMPILER BARRIER section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800901
Paul E. McKenneyebff09a2016-06-15 16:08:17 -0700902 (*) Control dependencies apply only to the then-clause and else-clause
903 of the if-statement containing the control dependency, including
904 any functions that these two clauses call. Control dependencies
905 do -not- apply to code following the if-statement containing the
906 control dependency.
907
Paul E. McKenneyff382812015-02-17 10:00:06 -0800908 (*) Control dependencies pair normally with other types of barriers.
909
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700910 (*) Control dependencies do -not- provide multicopy atomicity. If you
911 need all the CPUs to see a given store at the same time, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100912
Paul E. McKenneyc8241f82016-12-13 16:42:32 -0800913 (*) Compilers do not understand control dependencies. It is therefore
914 your job to ensure that they do not break your code.
915
David Howells108b42b2006-03-31 16:00:29 +0100916
917SMP BARRIER PAIRING
918-------------------
919
920When dealing with CPU-CPU interactions, certain types of memory barrier should
921always be paired. A lack of appropriate pairing is almost certainly an error.
922
Paul E. McKenneyff382812015-02-17 10:00:06 -0800923General barriers pair with each other, though they also pair with most
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -0700924other types of barriers, albeit without multicopy atomicity. An acquire
925barrier pairs with a release barrier, but both may also pair with other
926barriers, including of course general barriers. A write barrier pairs
927with a data dependency barrier, a control dependency, an acquire barrier,
928a release barrier, a read barrier, or a general barrier. Similarly a
929read barrier, control dependency, or a data dependency barrier pairs
930with a write barrier, an acquire barrier, a release barrier, or a
931general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100932
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800933 CPU 1 CPU 2
934 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700935 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100936 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700937 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800938 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700939 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100940
941Or:
942
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800943 CPU 1 CPU 2
944 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100945 a = 1;
946 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700947 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800948 <data dependency barrier>
949 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100950
Paul E. McKenneyff382812015-02-17 10:00:06 -0800951Or even:
952
953 CPU 1 CPU 2
954 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700955 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800956 <general barrier>
Scott Tsaid92f8422017-09-20 02:16:00 +0800957 WRITE_ONCE(x, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800958 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700959 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800960 }
961
962 assert(r1 == 0 || r2 == 0);
963
David Howells108b42b2006-03-31 16:00:29 +0100964Basically, the read barrier always has to be there, even though it can be of
965the "weaker" type.
966
David Howells670bd952006-06-10 09:54:12 -0700967[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700968match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700969versa:
970
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800971 CPU 1 CPU 2
972 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700973 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
974 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800975 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700976 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
977 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700978
David Howells108b42b2006-03-31 16:00:29 +0100979
980EXAMPLES OF MEMORY BARRIER SEQUENCES
981------------------------------------
982
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700983Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100984Consider the following sequence of events:
985
986 CPU 1
987 =======================
988 STORE A = 1
989 STORE B = 2
990 STORE C = 3
991 <write barrier>
992 STORE D = 4
993 STORE E = 5
994
995This sequence of events is committed to the memory coherence system in an order
996that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +0200997STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +0100998}:
999
1000 +-------+ : :
1001 | | +------+
1002 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001003 | | : +------+ }----- \ -----> Events perceptible to
1004 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +01001005 | | : +------+ }
1006 | CPU 1 | : | B=2 | }
1007 | | +------+ }
1008 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
1009 | | +------+ } requires all stores prior to the
1010 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001011 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +01001012 | |------>| D=4 | }
1013 | | +------+
1014 +-------+ : :
1015 |
David Howells670bd952006-06-10 09:54:12 -07001016 | Sequence in which stores are committed to the
1017 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +01001018 V
1019
1020
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001021Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +01001022loads. Consider the following sequence of events:
1023
1024 CPU 1 CPU 2
1025 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -07001026 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +01001027 STORE A = 1
1028 STORE B = 2
1029 <write barrier>
1030 STORE C = &B LOAD X
1031 STORE D = 4 LOAD C (gets &B)
1032 LOAD *C (reads B)
1033
1034Without intervention, CPU 2 may perceive the events on CPU 1 in some
1035effectively random order, despite the write barrier issued by CPU 1:
1036
1037 +-------+ : : : :
1038 | | +------+ +-------+ | Sequence of update
1039 | |------>| B=2 |----- --->| Y->8 | | of perception on
1040 | | : +------+ \ +-------+ | CPU 2
1041 | CPU 1 | : | A=1 | \ --->| C->&Y | V
1042 | | +------+ | +-------+
1043 | | wwwwwwwwwwwwwwww | : :
1044 | | +------+ | : :
1045 | | : | C=&B |--- | : : +-------+
1046 | | : +------+ \ | +-------+ | |
1047 | |------>| D=4 | ----------->| C->&B |------>| |
1048 | | +------+ | +-------+ | |
1049 +-------+ : : | : : | |
1050 | : : | |
1051 | : : | CPU 2 |
1052 | +-------+ | |
1053 Apparently incorrect ---> | | B->7 |------>| |
1054 perception of B (!) | +-------+ | |
1055 | : : | |
1056 | +-------+ | |
1057 The load of X holds ---> \ | X->9 |------>| |
1058 up the maintenance \ +-------+ | |
1059 of coherence of B ----->| B->2 | +-------+
1060 +-------+
1061 : :
1062
1063
1064In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +02001065(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +01001066
1067If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -07001068and the load of *C (ie: B) on CPU 2:
1069
1070 CPU 1 CPU 2
1071 ======================= =======================
1072 { B = 7; X = 9; Y = 8; C = &Y }
1073 STORE A = 1
1074 STORE B = 2
1075 <write barrier>
1076 STORE C = &B LOAD X
1077 STORE D = 4 LOAD C (gets &B)
1078 <data dependency barrier>
1079 LOAD *C (reads B)
1080
1081then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +01001082
1083 +-------+ : : : :
1084 | | +------+ +-------+
1085 | |------>| B=2 |----- --->| Y->8 |
1086 | | : +------+ \ +-------+
1087 | CPU 1 | : | A=1 | \ --->| C->&Y |
1088 | | +------+ | +-------+
1089 | | wwwwwwwwwwwwwwww | : :
1090 | | +------+ | : :
1091 | | : | C=&B |--- | : : +-------+
1092 | | : +------+ \ | +-------+ | |
1093 | |------>| D=4 | ----------->| C->&B |------>| |
1094 | | +------+ | +-------+ | |
1095 +-------+ : : | : : | |
1096 | : : | |
1097 | : : | CPU 2 |
1098 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001099 | | X->9 |------>| |
1100 | +-------+ | |
1101 Makes sure all effects ---> \ ddddddddddddddddd | |
1102 prior to the store of C \ +-------+ | |
1103 are perceptible to ----->| B->2 |------>| |
1104 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001105 : : +-------+
1106
1107
1108And thirdly, a read barrier acts as a partial order on loads. Consider the
1109following sequence of events:
1110
1111 CPU 1 CPU 2
1112 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001113 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001114 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001115 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001116 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001117 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001118 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001119
1120Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1121some effectively random order, despite the write barrier issued by CPU 1:
1122
David Howells670bd952006-06-10 09:54:12 -07001123 +-------+ : : : :
1124 | | +------+ +-------+
1125 | |------>| A=1 |------ --->| A->0 |
1126 | | +------+ \ +-------+
1127 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1128 | | +------+ | +-------+
1129 | |------>| B=2 |--- | : :
1130 | | +------+ \ | : : +-------+
1131 +-------+ : : \ | +-------+ | |
1132 ---------->| B->2 |------>| |
1133 | +-------+ | CPU 2 |
1134 | | A->0 |------>| |
1135 | +-------+ | |
1136 | : : +-------+
1137 \ : :
1138 \ +-------+
1139 ---->| A->1 |
1140 +-------+
1141 : :
David Howells108b42b2006-03-31 16:00:29 +01001142
1143
David Howells6bc39272006-06-25 05:49:22 -07001144If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001145load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001146
David Howells670bd952006-06-10 09:54:12 -07001147 CPU 1 CPU 2
1148 ======================= =======================
1149 { A = 0, B = 9 }
1150 STORE A=1
1151 <write barrier>
1152 STORE B=2
1153 LOAD B
1154 <read barrier>
1155 LOAD A
1156
1157then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
11582:
1159
1160 +-------+ : : : :
1161 | | +------+ +-------+
1162 | |------>| A=1 |------ --->| A->0 |
1163 | | +------+ \ +-------+
1164 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1165 | | +------+ | +-------+
1166 | |------>| B=2 |--- | : :
1167 | | +------+ \ | : : +-------+
1168 +-------+ : : \ | +-------+ | |
1169 ---------->| B->2 |------>| |
1170 | +-------+ | CPU 2 |
1171 | : : | |
1172 | : : | |
1173 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1174 barrier causes all effects \ +-------+ | |
1175 prior to the storage of B ---->| A->1 |------>| |
1176 to be perceptible to CPU 2 +-------+ | |
1177 : : +-------+
1178
1179
1180To illustrate this more completely, consider what could happen if the code
1181contained a load of A either side of the read barrier:
1182
1183 CPU 1 CPU 2
1184 ======================= =======================
1185 { A = 0, B = 9 }
1186 STORE A=1
1187 <write barrier>
1188 STORE B=2
1189 LOAD B
1190 LOAD A [first load of A]
1191 <read barrier>
1192 LOAD A [second load of A]
1193
1194Even though the two loads of A both occur after the load of B, they may both
1195come up with different values:
1196
1197 +-------+ : : : :
1198 | | +------+ +-------+
1199 | |------>| A=1 |------ --->| A->0 |
1200 | | +------+ \ +-------+
1201 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1202 | | +------+ | +-------+
1203 | |------>| B=2 |--- | : :
1204 | | +------+ \ | : : +-------+
1205 +-------+ : : \ | +-------+ | |
1206 ---------->| B->2 |------>| |
1207 | +-------+ | CPU 2 |
1208 | : : | |
1209 | : : | |
1210 | +-------+ | |
1211 | | A->0 |------>| 1st |
1212 | +-------+ | |
1213 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1214 barrier causes all effects \ +-------+ | |
1215 prior to the storage of B ---->| A->1 |------>| 2nd |
1216 to be perceptible to CPU 2 +-------+ | |
1217 : : +-------+
1218
1219
1220But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1221before the read barrier completes anyway:
1222
1223 +-------+ : : : :
1224 | | +------+ +-------+
1225 | |------>| A=1 |------ --->| A->0 |
1226 | | +------+ \ +-------+
1227 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1228 | | +------+ | +-------+
1229 | |------>| B=2 |--- | : :
1230 | | +------+ \ | : : +-------+
1231 +-------+ : : \ | +-------+ | |
1232 ---------->| B->2 |------>| |
1233 | +-------+ | CPU 2 |
1234 | : : | |
1235 \ : : | |
1236 \ +-------+ | |
1237 ---->| A->1 |------>| 1st |
1238 +-------+ | |
1239 rrrrrrrrrrrrrrrrr | |
1240 +-------+ | |
1241 | A->1 |------>| 2nd |
1242 +-------+ | |
1243 : : +-------+
1244
1245
1246The guarantee is that the second load will always come up with A == 1 if the
1247load of B came up with B == 2. No such guarantee exists for the first load of
1248A; that may come up with either A == 0 or A == 1.
1249
1250
1251READ MEMORY BARRIERS VS LOAD SPECULATION
1252----------------------------------------
1253
1254Many CPUs speculate with loads: that is they see that they will need to load an
1255item from memory, and they find a time where they're not using the bus for any
1256other loads, and so do the load in advance - even though they haven't actually
1257got to that point in the instruction execution flow yet. This permits the
1258actual load instruction to potentially complete immediately because the CPU
1259already has the value to hand.
1260
1261It may turn out that the CPU didn't actually need the value - perhaps because a
1262branch circumvented the load - in which case it can discard the value or just
1263cache it for later use.
1264
1265Consider:
1266
Ingo Molnare0edc782013-11-22 11:24:53 +01001267 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001268 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001269 LOAD B
1270 DIVIDE } Divide instructions generally
1271 DIVIDE } take a long time to perform
1272 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001273
1274Which might appear as this:
1275
1276 : : +-------+
1277 +-------+ | |
1278 --->| B->2 |------>| |
1279 +-------+ | CPU 2 |
1280 : :DIVIDE | |
1281 +-------+ | |
1282 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1283 division speculates on the +-------+ ~ | |
1284 LOAD of A : : ~ | |
1285 : :DIVIDE | |
1286 : : ~ | |
1287 Once the divisions are complete --> : : ~-->| |
1288 the CPU can then perform the : : | |
1289 LOAD with immediate effect : : +-------+
1290
1291
1292Placing a read barrier or a data dependency barrier just before the second
1293load:
1294
Ingo Molnare0edc782013-11-22 11:24:53 +01001295 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001296 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001297 LOAD B
1298 DIVIDE
1299 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001300 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001301 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001302
1303will force any value speculatively obtained to be reconsidered to an extent
1304dependent on the type of barrier used. If there was no change made to the
1305speculated memory location, then the speculated value will just be used:
1306
1307 : : +-------+
1308 +-------+ | |
1309 --->| B->2 |------>| |
1310 +-------+ | CPU 2 |
1311 : :DIVIDE | |
1312 +-------+ | |
1313 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1314 division speculates on the +-------+ ~ | |
1315 LOAD of A : : ~ | |
1316 : :DIVIDE | |
1317 : : ~ | |
1318 : : ~ | |
1319 rrrrrrrrrrrrrrrr~ | |
1320 : : ~ | |
1321 : : ~-->| |
1322 : : | |
1323 : : +-------+
1324
1325
1326but if there was an update or an invalidation from another CPU pending, then
1327the speculation will be cancelled and the value reloaded:
1328
1329 : : +-------+
1330 +-------+ | |
1331 --->| B->2 |------>| |
1332 +-------+ | CPU 2 |
1333 : :DIVIDE | |
1334 +-------+ | |
1335 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1336 division speculates on the +-------+ ~ | |
1337 LOAD of A : : ~ | |
1338 : :DIVIDE | |
1339 : : ~ | |
1340 : : ~ | |
1341 rrrrrrrrrrrrrrrrr | |
1342 +-------+ | |
1343 The speculation is discarded ---> --->| A->1 |------>| |
1344 and an updated value is +-------+ | |
1345 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001346
1347
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001348MULTICOPY ATOMICITY
1349--------------------
Paul E. McKenney241e6662011-02-10 16:54:50 -08001350
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001351Multicopy atomicity is a deeply intuitive notion about ordering that is
1352not always provided by real computer systems, namely that a given store
Alan Stern0902b1f2017-09-01 07:53:34 -07001353becomes visible at the same time to all CPUs, or, alternatively, that all
1354CPUs agree on the order in which all stores become visible. However,
1355support of full multicopy atomicity would rule out valuable hardware
1356optimizations, so a weaker form called ``other multicopy atomicity''
1357instead guarantees only that a given store becomes visible at the same
1358time to all -other- CPUs. The remainder of this document discusses this
1359weaker form, but for brevity will call it simply ``multicopy atomicity''.
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001360
1361The following example demonstrates multicopy atomicity:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001362
1363 CPU 1 CPU 2 CPU 3
1364 ======================= ======================= =======================
1365 { X = 0, Y = 0 }
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001366 STORE X=1 r1=LOAD X (reads 1) LOAD Y (reads 1)
1367 <general barrier> <read barrier>
1368 STORE Y=r1 LOAD X
Paul E. McKenney241e6662011-02-10 16:54:50 -08001369
Alan Stern0902b1f2017-09-01 07:53:34 -07001370Suppose that CPU 2's load from X returns 1, which it then stores to Y,
1371and CPU 3's load from Y returns 1. This indicates that CPU 1's store
1372to X precedes CPU 2's load from X and that CPU 2's store to Y precedes
1373CPU 3's load from Y. In addition, the memory barriers guarantee that
1374CPU 2 executes its load before its store, and CPU 3 loads from Y before
1375it loads from X. The question is then "Can CPU 3's load from X return 0?"
Paul E. McKenney241e6662011-02-10 16:54:50 -08001376
Alan Stern0902b1f2017-09-01 07:53:34 -07001377Because CPU 3's load from X in some sense comes after CPU 2's load, it
Paul E. McKenney241e6662011-02-10 16:54:50 -08001378is natural to expect that CPU 3's load from X must therefore return 1.
Alan Stern0902b1f2017-09-01 07:53:34 -07001379This expectation follows from multicopy atomicity: if a load executing
1380on CPU B follows a load from the same variable executing on CPU A (and
1381CPU A did not originally store the value which it read), then on
1382multicopy-atomic systems, CPU B's load must return either the same value
1383that CPU A's load did or some later value. However, the Linux kernel
1384does not require systems to be multicopy atomic.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001385
Alan Stern0902b1f2017-09-01 07:53:34 -07001386The use of a general memory barrier in the example above compensates
1387for any lack of multicopy atomicity. In the example, if CPU 2's load
1388from X returns 1 and CPU 3's load from Y returns 1, then CPU 3's load
1389from X must indeed also return 1.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001390
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001391However, dependencies, read barriers, and write barriers are not always
1392able to compensate for non-multicopy atomicity. For example, suppose
1393that CPU 2's general barrier is removed from the above example, leaving
1394only the data dependency shown below:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001395
1396 CPU 1 CPU 2 CPU 3
1397 ======================= ======================= =======================
1398 { X = 0, Y = 0 }
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001399 STORE X=1 r1=LOAD X (reads 1) LOAD Y (reads 1)
1400 <data dependency> <read barrier>
1401 STORE Y=r1 LOAD X (reads 0)
Paul E. McKenney241e6662011-02-10 16:54:50 -08001402
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001403This substitution allows non-multicopy atomicity to run rampant: in
1404this example, it is perfectly legal for CPU 2's load from X to return 1,
1405CPU 3's load from Y to return 1, and its load from X to return 0.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001406
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001407The key point is that although CPU 2's data dependency orders its load
Alan Stern0902b1f2017-09-01 07:53:34 -07001408and store, it does not guarantee to order CPU 1's store. Thus, if this
1409example runs on a non-multicopy-atomic system where CPUs 1 and 2 share a
1410store buffer or a level of cache, CPU 2 might have early access to CPU 1's
1411writes. General barriers are therefore required to ensure that all CPUs
1412agree on the combined order of multiple accesses.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001413
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001414General barriers can compensate not only for non-multicopy atomicity,
1415but can also generate additional ordering that can ensure that -all-
1416CPUs will perceive the same order of -all- operations. In contrast, a
1417chain of release-acquire pairs do not provide this additional ordering,
1418which means that only those CPUs on the chain are guaranteed to agree
1419on the combined order of the accesses. For example, switching to C code
1420in deference to the ghost of Herman Hollerith:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001421
1422 int u, v, x, y, z;
1423
1424 void cpu0(void)
1425 {
1426 r0 = smp_load_acquire(&x);
1427 WRITE_ONCE(u, 1);
1428 smp_store_release(&y, 1);
1429 }
1430
1431 void cpu1(void)
1432 {
1433 r1 = smp_load_acquire(&y);
1434 r4 = READ_ONCE(v);
1435 r5 = READ_ONCE(u);
1436 smp_store_release(&z, 1);
1437 }
1438
1439 void cpu2(void)
1440 {
1441 r2 = smp_load_acquire(&z);
1442 smp_store_release(&x, 1);
1443 }
1444
1445 void cpu3(void)
1446 {
1447 WRITE_ONCE(v, 1);
1448 smp_mb();
1449 r3 = READ_ONCE(u);
1450 }
1451
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001452Because cpu0(), cpu1(), and cpu2() participate in a chain of
1453smp_store_release()/smp_load_acquire() pairs, the following outcome
1454is prohibited:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001455
1456 r0 == 1 && r1 == 1 && r2 == 1
1457
1458Furthermore, because of the release-acquire relationship between cpu0()
1459and cpu1(), cpu1() must see cpu0()'s writes, so that the following
1460outcome is prohibited:
1461
1462 r1 == 1 && r5 == 0
1463
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001464However, the ordering provided by a release-acquire chain is local
1465to the CPUs participating in that chain and does not apply to cpu3(),
1466at least aside from stores. Therefore, the following outcome is possible:
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001467
1468 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
1469
Paul E. McKenney37ef0342016-01-25 22:12:34 -08001470As an aside, the following outcome is also possible:
1471
1472 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
1473
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001474Although cpu0(), cpu1(), and cpu2() will see their respective reads and
1475writes in order, CPUs not involved in the release-acquire chain might
1476well disagree on the order. This disagreement stems from the fact that
1477the weak memory-barrier instructions used to implement smp_load_acquire()
1478and smp_store_release() are not required to order prior stores against
1479subsequent loads in all cases. This means that cpu3() can see cpu0()'s
1480store to u as happening -after- cpu1()'s load from v, even though
1481both cpu0() and cpu1() agree that these two operations occurred in the
1482intended order.
1483
1484However, please keep in mind that smp_load_acquire() is not magic.
1485In particular, it simply reads from its argument with ordering. It does
1486-not- ensure that any particular value will be read. Therefore, the
1487following outcome is possible:
1488
1489 r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
1490
1491Note that this outcome can happen even on a mythical sequentially
1492consistent system where nothing is ever reordered.
1493
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07001494To reiterate, if your code requires full ordering of all operations,
1495use general barriers throughout.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001496
1497
David Howells108b42b2006-03-31 16:00:29 +01001498========================
1499EXPLICIT KERNEL BARRIERS
1500========================
1501
1502The Linux kernel has a variety of different barriers that act at different
1503levels:
1504
1505 (*) Compiler barrier.
1506
1507 (*) CPU memory barriers.
1508
1509 (*) MMIO write barrier.
1510
1511
1512COMPILER BARRIER
1513----------------
1514
1515The Linux kernel has an explicit compiler barrier function that prevents the
1516compiler from moving the memory accesses either side of it to the other side:
1517
1518 barrier();
1519
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001520This is a general barrier -- there are no read-read or write-write
1521variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1522thought of as weak forms of barrier() that affect only the specific
1523accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001524
Paul E. McKenney692118d2013-12-11 13:59:07 -08001525The barrier() function has the following effects:
1526
1527 (*) Prevents the compiler from reordering accesses following the
1528 barrier() to precede any accesses preceding the barrier().
1529 One example use for this property is to ease communication between
1530 interrupt-handler code and the code that was interrupted.
1531
1532 (*) Within a loop, forces the compiler to load the variables used
1533 in that loop's conditional on each pass through that loop.
1534
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001535The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1536optimizations that, while perfectly safe in single-threaded code, can
1537be fatal in concurrent code. Here are some examples of these sorts
1538of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001539
Paul E. McKenney449f7412014-01-02 15:03:50 -08001540 (*) The compiler is within its rights to reorder loads and stores
1541 to the same variable, and in some cases, the CPU is within its
1542 rights to reorder loads to the same variable. This means that
1543 the following code:
1544
1545 a[0] = x;
1546 a[1] = x;
1547
1548 Might result in an older value of x stored in a[1] than in a[0].
1549 Prevent both the compiler and the CPU from doing this as follows:
1550
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001551 a[0] = READ_ONCE(x);
1552 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001553
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001554 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1555 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001556
Paul E. McKenney692118d2013-12-11 13:59:07 -08001557 (*) The compiler is within its rights to merge successive loads from
1558 the same variable. Such merging can cause the compiler to "optimize"
1559 the following code:
1560
1561 while (tmp = a)
1562 do_something_with(tmp);
1563
1564 into the following code, which, although in some sense legitimate
1565 for single-threaded code, is almost certainly not what the developer
1566 intended:
1567
1568 if (tmp = a)
1569 for (;;)
1570 do_something_with(tmp);
1571
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001572 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001573
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001574 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001575 do_something_with(tmp);
1576
1577 (*) The compiler is within its rights to reload a variable, for example,
1578 in cases where high register pressure prevents the compiler from
1579 keeping all data of interest in registers. The compiler might
1580 therefore optimize the variable 'tmp' out of our previous example:
1581
1582 while (tmp = a)
1583 do_something_with(tmp);
1584
1585 This could result in the following code, which is perfectly safe in
1586 single-threaded code, but can be fatal in concurrent code:
1587
1588 while (a)
1589 do_something_with(a);
1590
1591 For example, the optimized version of this code could result in
1592 passing a zero to do_something_with() in the case where the variable
1593 a was modified by some other CPU between the "while" statement and
1594 the call to do_something_with().
1595
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001596 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001597
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001598 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001599 do_something_with(tmp);
1600
1601 Note that if the compiler runs short of registers, it might save
1602 tmp onto the stack. The overhead of this saving and later restoring
1603 is why compilers reload variables. Doing so is perfectly safe for
1604 single-threaded code, so you need to tell the compiler about cases
1605 where it is not safe.
1606
1607 (*) The compiler is within its rights to omit a load entirely if it knows
1608 what the value will be. For example, if the compiler can prove that
1609 the value of variable 'a' is always zero, it can optimize this code:
1610
1611 while (tmp = a)
1612 do_something_with(tmp);
1613
1614 Into this:
1615
1616 do { } while (0);
1617
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001618 This transformation is a win for single-threaded code because it
1619 gets rid of a load and a branch. The problem is that the compiler
1620 will carry out its proof assuming that the current CPU is the only
1621 one updating variable 'a'. If variable 'a' is shared, then the
1622 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1623 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001624
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001625 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001626 do_something_with(tmp);
1627
1628 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001629 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001630 do the following and MAX is a preprocessor macro with the value 1:
1631
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001632 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001633 do_something_with(tmp);
1634
1635 Then the compiler knows that the result of the "%" operator applied
1636 to MAX will always be zero, again allowing the compiler to optimize
1637 the code into near-nonexistence. (It will still load from the
1638 variable 'a'.)
1639
1640 (*) Similarly, the compiler is within its rights to omit a store entirely
1641 if it knows that the variable already has the value being stored.
1642 Again, the compiler assumes that the current CPU is the only one
1643 storing into the variable, which can cause the compiler to do the
1644 wrong thing for shared variables. For example, suppose you have
1645 the following:
1646
1647 a = 0;
SeongJae Park65f95ff2016-02-22 08:28:29 -08001648 ... Code that does not store to variable a ...
Paul E. McKenney692118d2013-12-11 13:59:07 -08001649 a = 0;
1650
1651 The compiler sees that the value of variable 'a' is already zero, so
1652 it might well omit the second store. This would come as a fatal
1653 surprise if some other CPU might have stored to variable 'a' in the
1654 meantime.
1655
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001656 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001657 wrong guess:
1658
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001659 WRITE_ONCE(a, 0);
SeongJae Park65f95ff2016-02-22 08:28:29 -08001660 ... Code that does not store to variable a ...
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001661 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001662
1663 (*) The compiler is within its rights to reorder memory accesses unless
1664 you tell it not to. For example, consider the following interaction
1665 between process-level code and an interrupt handler:
1666
1667 void process_level(void)
1668 {
1669 msg = get_message();
1670 flag = true;
1671 }
1672
1673 void interrupt_handler(void)
1674 {
1675 if (flag)
1676 process_message(msg);
1677 }
1678
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001679 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001680 process_level() to the following, in fact, this might well be a
1681 win for single-threaded code:
1682
1683 void process_level(void)
1684 {
1685 flag = true;
1686 msg = get_message();
1687 }
1688
1689 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001690 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001691 to prevent this as follows:
1692
1693 void process_level(void)
1694 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001695 WRITE_ONCE(msg, get_message());
1696 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001697 }
1698
1699 void interrupt_handler(void)
1700 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001701 if (READ_ONCE(flag))
1702 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001703 }
1704
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001705 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1706 interrupt_handler() are needed if this interrupt handler can itself
1707 be interrupted by something that also accesses 'flag' and 'msg',
1708 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1709 and WRITE_ONCE() are not needed in interrupt_handler() other than
1710 for documentation purposes. (Note also that nested interrupts
1711 do not typically occur in modern Linux kernels, in fact, if an
1712 interrupt handler returns with interrupts enabled, you will get a
1713 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001714
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001715 You should assume that the compiler can move READ_ONCE() and
1716 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1717 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001718
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001719 This effect could also be achieved using barrier(), but READ_ONCE()
1720 and WRITE_ONCE() are more selective: With READ_ONCE() and
1721 WRITE_ONCE(), the compiler need only forget the contents of the
1722 indicated memory locations, while with barrier() the compiler must
1723 discard the value of all memory locations that it has currented
1724 cached in any machine registers. Of course, the compiler must also
1725 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1726 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001727
1728 (*) The compiler is within its rights to invent stores to a variable,
1729 as in the following example:
1730
1731 if (a)
1732 b = a;
1733 else
1734 b = 42;
1735
1736 The compiler might save a branch by optimizing this as follows:
1737
1738 b = 42;
1739 if (a)
1740 b = a;
1741
1742 In single-threaded code, this is not only safe, but also saves
1743 a branch. Unfortunately, in concurrent code, this optimization
1744 could cause some other CPU to see a spurious value of 42 -- even
1745 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001746 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001747
1748 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001749 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001750 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001751 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001752
1753 The compiler can also invent loads. These are usually less
1754 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001755 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001756 invented loads.
1757
1758 (*) For aligned memory locations whose size allows them to be accessed
1759 with a single memory-reference instruction, prevents "load tearing"
1760 and "store tearing," in which a single large access is replaced by
1761 multiple smaller accesses. For example, given an architecture having
1762 16-bit store instructions with 7-bit immediate fields, the compiler
1763 might be tempted to use two 16-bit store-immediate instructions to
1764 implement the following 32-bit store:
1765
1766 p = 0x00010002;
1767
1768 Please note that GCC really does use this sort of optimization,
1769 which is not surprising given that it would likely take more
1770 than two instructions to build the constant and then store it.
1771 This optimization can therefore be a win in single-threaded code.
1772 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1773 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001774 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001775
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001776 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001777
1778 Use of packed structures can also result in load and store tearing,
1779 as in this example:
1780
1781 struct __attribute__((__packed__)) foo {
1782 short a;
1783 int b;
1784 short c;
1785 };
1786 struct foo foo1, foo2;
1787 ...
1788
1789 foo2.a = foo1.a;
1790 foo2.b = foo1.b;
1791 foo2.c = foo1.c;
1792
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001793 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1794 volatile markings, the compiler would be well within its rights to
1795 implement these three assignment statements as a pair of 32-bit
1796 loads followed by a pair of 32-bit stores. This would result in
1797 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1798 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001799
1800 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001801 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001802 foo2.c = foo1.c;
1803
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001804All that aside, it is never necessary to use READ_ONCE() and
1805WRITE_ONCE() on a variable that has been marked volatile. For example,
1806because 'jiffies' is marked volatile, it is never necessary to
1807say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1808WRITE_ONCE() are implemented as volatile casts, which has no effect when
1809its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001810
1811Please note that these compiler barriers have no direct effect on the CPU,
1812which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001813
1814
1815CPU MEMORY BARRIERS
1816-------------------
1817
1818The Linux kernel has eight basic CPU memory barriers:
1819
1820 TYPE MANDATORY SMP CONDITIONAL
1821 =============== ======================= ===========================
1822 GENERAL mb() smp_mb()
1823 WRITE wmb() smp_wmb()
1824 READ rmb() smp_rmb()
Paul E. McKenney9ad3c142017-11-27 09:20:40 -08001825 DATA DEPENDENCY READ_ONCE()
David Howells108b42b2006-03-31 16:00:29 +01001826
1827
Nick Piggin73f10282008-05-14 06:35:11 +02001828All memory barriers except the data dependency barriers imply a compiler
SeongJae Park0b6fa342016-04-12 08:52:53 -07001829barrier. Data dependencies do not impose any additional compiler ordering.
Nick Piggin73f10282008-05-14 06:35:11 +02001830
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001831Aside: In the case of data dependencies, the compiler would be expected
1832to issue the loads in the correct order (eg. `a[b]` would have to load
1833the value of b before loading a[b]), however there is no guarantee in
1834the C specification that the compiler may not speculate the value of b
1835(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
SeongJae Park0b6fa342016-04-12 08:52:53 -07001836tmp = a[b]; ). There is also the problem of a compiler reloading b after
1837having loaded a[b], thus having a newer copy of b than a[b]. A consensus
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001838has not yet been reached about these problems, however the READ_ONCE()
1839macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001840
1841SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001842systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001843and will order overlapping accesses correctly with respect to itself.
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001844However, see the subsection on "Virtual Machine Guests" below.
David Howells108b42b2006-03-31 16:00:29 +01001845
1846[!] Note that SMP memory barriers _must_ be used to control the ordering of
1847references to shared memory on SMP systems, though the use of locking instead
1848is sufficient.
1849
1850Mandatory barriers should not be used to control SMP effects, since mandatory
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001851barriers impose unnecessary overhead on both SMP and UP systems. They may,
1852however, be used to control MMIO effects on accesses through relaxed memory I/O
1853windows. These barriers are required even on non-SMP systems as they affect
1854the order in which memory operations appear to a device by prohibiting both the
1855compiler and the CPU from reordering them.
David Howells108b42b2006-03-31 16:00:29 +01001856
1857
1858There are some more advanced barrier functions:
1859
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001860 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001861
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001862 This assigns the value to the variable and then inserts a full memory
Davidlohr Bueso2d142e52015-10-27 12:53:51 -07001863 barrier after it. It isn't guaranteed to insert anything more than a
1864 compiler barrier in a UP compilation.
David Howells108b42b2006-03-31 16:00:29 +01001865
1866
Peter Zijlstra1b156112014-03-13 19:00:35 +01001867 (*) smp_mb__before_atomic();
1868 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001869
Peter Zijlstra1b156112014-03-13 19:00:35 +01001870 These are for use with atomic (such as add, subtract, increment and
1871 decrement) functions that don't return a value, especially when used for
1872 reference counting. These functions do not imply memory barriers.
1873
1874 These are also used for atomic bitop functions that do not return a
1875 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001876
1877 As an example, consider a piece of code that marks an object as being dead
1878 and then decrements the object's reference count:
1879
1880 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001881 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001882 atomic_dec(&obj->ref_count);
1883
1884 This makes sure that the death mark on the object is perceived to be set
1885 *before* the reference counter is decremented.
1886
Peter Zijlstra706eeb32017-06-12 14:50:27 +02001887 See Documentation/atomic_{t,bitops}.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01001888
1889
Alexander Duyck1077fa32014-12-11 15:02:06 -08001890 (*) dma_wmb();
1891 (*) dma_rmb();
1892
1893 These are for use with consistent memory to guarantee the ordering
1894 of writes or reads of shared memory accessible to both the CPU and a
1895 DMA capable device.
1896
1897 For example, consider a device driver that shares memory with a device
1898 and uses a descriptor status value to indicate if the descriptor belongs
1899 to the device or the CPU, and a doorbell to notify it when new
1900 descriptors are available:
1901
1902 if (desc->status != DEVICE_OWN) {
1903 /* do not read data until we own descriptor */
1904 dma_rmb();
1905
1906 /* read/modify data */
1907 read_data = desc->data;
1908 desc->data = write_data;
1909
1910 /* flush modifications before status update */
1911 dma_wmb();
1912
1913 /* assign ownership */
1914 desc->status = DEVICE_OWN;
1915
1916 /* force memory to sync before notifying device via MMIO */
1917 wmb();
1918
1919 /* notify device of new descriptors */
1920 writel(DESC_NOTIFY, doorbell);
1921 }
1922
1923 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001924 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001925 us to guarantee the data is written to the descriptor before the device
1926 can see it now has ownership. The wmb() is needed to guarantee that the
1927 cache coherent memory writes have completed before attempting a write to
1928 the cache incoherent MMIO region.
1929
1930 See Documentation/DMA-API.txt for more information on consistent memory.
1931
SeongJae Parkdfeccea2016-08-11 11:17:40 -07001932
David Howells108b42b2006-03-31 16:00:29 +01001933MMIO WRITE BARRIER
1934------------------
1935
1936The Linux kernel also has a special barrier for use with memory-mapped I/O
1937writes:
1938
1939 mmiowb();
1940
1941This is a variation on the mandatory write barrier that causes writes to weakly
1942ordered I/O regions to be partially ordered. Its effects may go beyond the
1943CPU->Hardware interface and actually affect the hardware at some level.
1944
SeongJae Park166bda72016-04-12 08:52:50 -07001945See the subsection "Acquires vs I/O accesses" for more information.
David Howells108b42b2006-03-31 16:00:29 +01001946
1947
1948===============================
1949IMPLICIT KERNEL MEMORY BARRIERS
1950===============================
1951
1952Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001953which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001954
1955This specification is a _minimum_ guarantee; any particular architecture may
1956provide more substantial guarantees, but these may not be relied upon outside
1957of arch specific code.
1958
1959
SeongJae Park166bda72016-04-12 08:52:50 -07001960LOCK ACQUISITION FUNCTIONS
1961--------------------------
David Howells108b42b2006-03-31 16:00:29 +01001962
1963The Linux kernel has a number of locking constructs:
1964
1965 (*) spin locks
1966 (*) R/W spin locks
1967 (*) mutexes
1968 (*) semaphores
1969 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001970
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001971In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001972for each construct. These operations all imply certain barriers:
1973
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001974 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001975
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001976 Memory operations issued after the ACQUIRE will be completed after the
1977 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001978
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001979 Memory operations issued before the ACQUIRE may be completed after
Peter Zijlstraa9668cd2017-06-07 17:51:27 +02001980 the ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001981
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001982 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001983
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001984 Memory operations issued before the RELEASE will be completed before the
1985 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001986
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001987 Memory operations issued after the RELEASE may be completed before the
1988 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001989
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001990 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001991
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001992 All ACQUIRE operations issued before another ACQUIRE operation will be
1993 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01001994
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001995 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01001996
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001997 All ACQUIRE operations issued before a RELEASE operation will be
1998 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01001999
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002000 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01002001
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002002 Certain locking variants of the ACQUIRE operation may fail, either due to
2003 being unable to get the lock immediately, or due to receiving an unblocked
David Howells108b42b2006-03-31 16:00:29 +01002004 signal whilst asleep waiting for the lock to become available. Failed
2005 locks do not imply any sort of barrier.
2006
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002007[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
2008one-way barriers is that the effects of instructions outside of a critical
2009section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01002010
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002011An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
2012because it is possible for an access preceding the ACQUIRE to happen after the
2013ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
2014the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07002015
2016 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002017 ACQUIRE M
2018 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07002019 *B = b;
2020
2021may occur as:
2022
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002023 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002024
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002025When the ACQUIRE and RELEASE are a lock acquisition and release,
2026respectively, this same reordering can occur if the lock's ACQUIRE and
2027RELEASE are to the same lock variable, but only from the perspective of
2028another CPU not holding that lock. In short, a ACQUIRE followed by an
2029RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002030
Paul E. McKenney12d560f2015-07-14 18:35:23 -07002031Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
2032not imply a full memory barrier. Therefore, the CPU's execution of the
2033critical sections corresponding to the RELEASE and the ACQUIRE can cross,
2034so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002035
2036 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002037 RELEASE M
2038 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002039 *B = b;
2040
2041could occur as:
2042
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002043 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08002044
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08002045It might appear that this reordering could introduce a deadlock.
2046However, this cannot happen because if such a deadlock threatened,
2047the RELEASE would simply complete, thereby avoiding the deadlock.
2048
2049 Why does this work?
2050
2051 One key point is that we are only talking about the CPU doing
2052 the reordering, not the compiler. If the compiler (or, for
2053 that matter, the developer) switched the operations, deadlock
2054 -could- occur.
2055
2056 But suppose the CPU reordered the operations. In this case,
2057 the unlock precedes the lock in the assembly code. The CPU
2058 simply elected to try executing the later lock operation first.
2059 If there is a deadlock, this lock operation will simply spin (or
2060 try to sleep, but more on that later). The CPU will eventually
2061 execute the unlock operation (which preceded the lock operation
2062 in the assembly code), which will unravel the potential deadlock,
2063 allowing the lock operation to succeed.
2064
2065 But what if the lock is a sleeplock? In that case, the code will
2066 try to enter the scheduler, where it will eventually encounter
2067 a memory barrier, which will force the earlier unlock operation
2068 to complete, again unraveling the deadlock. There might be
2069 a sleep-unlock race, but the locking primitive needs to resolve
2070 such races properly in any case.
2071
David Howells108b42b2006-03-31 16:00:29 +01002072Locks and semaphores may not provide any guarantee of ordering on UP compiled
2073systems, and so cannot be counted on in such a situation to actually achieve
2074anything at all - especially with respect to I/O accesses - unless combined
2075with interrupt disabling operations.
2076
SeongJae Parkd7cab362016-08-11 11:17:41 -07002077See also the section on "Inter-CPU acquiring barrier effects".
David Howells108b42b2006-03-31 16:00:29 +01002078
2079
2080As an example, consider the following:
2081
2082 *A = a;
2083 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002084 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01002085 *C = c;
2086 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002087 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002088 *E = e;
2089 *F = f;
2090
2091The following sequence of events is acceptable:
2092
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002093 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002094
2095 [+] Note that {*F,*A} indicates a combined access.
2096
2097But none of the following are:
2098
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002099 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
2100 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
2101 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
2102 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01002103
2104
2105
2106INTERRUPT DISABLING FUNCTIONS
2107-----------------------------
2108
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002109Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
2110(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01002111barriers are required in such a situation, they must be provided from some
2112other means.
2113
2114
David Howells50fa6102009-04-28 15:01:38 +01002115SLEEP AND WAKE-UP FUNCTIONS
2116---------------------------
2117
2118Sleeping and waking on an event flagged in global data can be viewed as an
2119interaction between two pieces of data: the task state of the task waiting for
2120the event and the global data used to indicate the event. To make sure that
2121these appear to happen in the right order, the primitives to begin the process
2122of going to sleep, and the primitives to initiate a wake up imply certain
2123barriers.
2124
2125Firstly, the sleeper normally follows something like this sequence of events:
2126
2127 for (;;) {
2128 set_current_state(TASK_UNINTERRUPTIBLE);
2129 if (event_indicated)
2130 break;
2131 schedule();
2132 }
2133
2134A general memory barrier is interpolated automatically by set_current_state()
2135after it has altered the task state:
2136
2137 CPU 1
2138 ===============================
2139 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002140 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01002141 STORE current->state
2142 <general barrier>
2143 LOAD event_indicated
2144
2145set_current_state() may be wrapped by:
2146
2147 prepare_to_wait();
2148 prepare_to_wait_exclusive();
2149
2150which therefore also imply a general memory barrier after setting the state.
2151The whole sequence above is available in various canned forms, all of which
2152interpolate the memory barrier in the right place:
2153
2154 wait_event();
2155 wait_event_interruptible();
2156 wait_event_interruptible_exclusive();
2157 wait_event_interruptible_timeout();
2158 wait_event_killable();
2159 wait_event_timeout();
2160 wait_on_bit();
2161 wait_on_bit_lock();
2162
2163
2164Secondly, code that performs a wake up normally follows something like this:
2165
2166 event_indicated = 1;
2167 wake_up(&event_wait_queue);
2168
2169or:
2170
2171 event_indicated = 1;
2172 wake_up_process(event_daemon);
2173
SeongJae Park0b6fa342016-04-12 08:52:53 -07002174A write memory barrier is implied by wake_up() and co. if and only if they
2175wake something up. The barrier occurs before the task state is cleared, and so
2176sits between the STORE to indicate the event and the STORE to set TASK_RUNNING:
David Howells50fa6102009-04-28 15:01:38 +01002177
2178 CPU 1 CPU 2
2179 =============================== ===============================
2180 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002181 smp_store_mb(); wake_up();
David Howells50fa6102009-04-28 15:01:38 +01002182 STORE current->state <write barrier>
2183 <general barrier> STORE current->state
2184 LOAD event_indicated
2185
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002186To repeat, this write memory barrier is present if and only if something
2187is actually awakened. To see this, consider the following sequence of
2188events, where X and Y are both initially zero:
2189
2190 CPU 1 CPU 2
2191 =============================== ===============================
2192 X = 1; STORE event_indicated
2193 smp_mb(); wake_up();
2194 Y = 1; wait_event(wq, Y == 1);
2195 wake_up(); load from Y sees 1, no memory barrier
2196 load from X might see 0
2197
2198In contrast, if a wakeup does occur, CPU 2's load from X would be guaranteed
2199to see 1.
2200
David Howells50fa6102009-04-28 15:01:38 +01002201The available waker functions include:
2202
2203 complete();
2204 wake_up();
2205 wake_up_all();
2206 wake_up_bit();
2207 wake_up_interruptible();
2208 wake_up_interruptible_all();
2209 wake_up_interruptible_nr();
2210 wake_up_interruptible_poll();
2211 wake_up_interruptible_sync();
2212 wake_up_interruptible_sync_poll();
2213 wake_up_locked();
2214 wake_up_locked_poll();
2215 wake_up_nr();
2216 wake_up_poll();
2217 wake_up_process();
2218
2219
2220[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2221order multiple stores before the wake-up with respect to loads of those stored
2222values after the sleeper has called set_current_state(). For instance, if the
2223sleeper does:
2224
2225 set_current_state(TASK_INTERRUPTIBLE);
2226 if (event_indicated)
2227 break;
2228 __set_current_state(TASK_RUNNING);
2229 do_something(my_data);
2230
2231and the waker does:
2232
2233 my_data = value;
2234 event_indicated = 1;
2235 wake_up(&event_wait_queue);
2236
2237there's no guarantee that the change to event_indicated will be perceived by
2238the sleeper as coming after the change to my_data. In such a circumstance, the
2239code on both sides must interpolate its own memory barriers between the
2240separate data accesses. Thus the above sleeper ought to do:
2241
2242 set_current_state(TASK_INTERRUPTIBLE);
2243 if (event_indicated) {
2244 smp_rmb();
2245 do_something(my_data);
2246 }
2247
2248and the waker should do:
2249
2250 my_data = value;
2251 smp_wmb();
2252 event_indicated = 1;
2253 wake_up(&event_wait_queue);
2254
2255
David Howells108b42b2006-03-31 16:00:29 +01002256MISCELLANEOUS FUNCTIONS
2257-----------------------
2258
2259Other functions that imply barriers:
2260
2261 (*) schedule() and similar imply full memory barriers.
2262
David Howells108b42b2006-03-31 16:00:29 +01002263
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002264===================================
2265INTER-CPU ACQUIRING BARRIER EFFECTS
2266===================================
David Howells108b42b2006-03-31 16:00:29 +01002267
2268On SMP systems locking primitives give a more substantial form of barrier: one
2269that does affect memory access ordering on other CPUs, within the context of
2270conflict on any particular lock.
2271
2272
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002273ACQUIRES VS MEMORY ACCESSES
2274---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002275
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002276Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002277three CPUs; then should the following sequence of events occur:
2278
2279 CPU 1 CPU 2
2280 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002281 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002282 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002283 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2284 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002285 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002286 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002287
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002288Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002289through *H occur in, other than the constraints imposed by the separate locks
SeongJae Park0b6fa342016-04-12 08:52:53 -07002290on the separate CPUs. It might, for example, see:
David Howells108b42b2006-03-31 16:00:29 +01002291
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002292 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002293
2294But it won't see any of:
2295
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002296 *B, *C or *D preceding ACQUIRE M
2297 *A, *B or *C following RELEASE M
2298 *F, *G or *H preceding ACQUIRE Q
2299 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002300
2301
David Howells108b42b2006-03-31 16:00:29 +01002302
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002303ACQUIRES VS I/O ACCESSES
2304------------------------
David Howells108b42b2006-03-31 16:00:29 +01002305
2306Under certain circumstances (especially involving NUMA), I/O accesses within
2307two spinlocked sections on two different CPUs may be seen as interleaved by the
2308PCI bridge, because the PCI bridge does not necessarily participate in the
2309cache-coherence protocol, and is therefore incapable of issuing the required
2310read memory barriers.
2311
2312For example:
2313
2314 CPU 1 CPU 2
2315 =============================== ===============================
2316 spin_lock(Q)
2317 writel(0, ADDR)
2318 writel(1, DATA);
2319 spin_unlock(Q);
2320 spin_lock(Q);
2321 writel(4, ADDR);
2322 writel(5, DATA);
2323 spin_unlock(Q);
2324
2325may be seen by the PCI bridge as follows:
2326
2327 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2328
2329which would probably cause the hardware to malfunction.
2330
2331
2332What is necessary here is to intervene with an mmiowb() before dropping the
2333spinlock, for example:
2334
2335 CPU 1 CPU 2
2336 =============================== ===============================
2337 spin_lock(Q)
2338 writel(0, ADDR)
2339 writel(1, DATA);
2340 mmiowb();
2341 spin_unlock(Q);
2342 spin_lock(Q);
2343 writel(4, ADDR);
2344 writel(5, DATA);
2345 mmiowb();
2346 spin_unlock(Q);
2347
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002348this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2349before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002350
2351
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002352Furthermore, following a store by a load from the same device obviates the need
2353for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002354is performed:
2355
2356 CPU 1 CPU 2
2357 =============================== ===============================
2358 spin_lock(Q)
2359 writel(0, ADDR)
2360 a = readl(DATA);
2361 spin_unlock(Q);
2362 spin_lock(Q);
2363 writel(4, ADDR);
2364 b = readl(DATA);
2365 spin_unlock(Q);
2366
2367
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002368See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002369
2370
2371=================================
2372WHERE ARE MEMORY BARRIERS NEEDED?
2373=================================
2374
2375Under normal operation, memory operation reordering is generally not going to
2376be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002377work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002378circumstances in which reordering definitely _could_ be a problem:
2379
2380 (*) Interprocessor interaction.
2381
2382 (*) Atomic operations.
2383
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002384 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002385
2386 (*) Interrupts.
2387
2388
2389INTERPROCESSOR INTERACTION
2390--------------------------
2391
2392When there's a system with more than one processor, more than one CPU in the
2393system may be working on the same data set at the same time. This can cause
2394synchronisation problems, and the usual way of dealing with them is to use
2395locks. Locks, however, are quite expensive, and so it may be preferable to
2396operate without the use of a lock if at all possible. In such a case
2397operations that affect both CPUs may have to be carefully ordered to prevent
2398a malfunction.
2399
2400Consider, for example, the R/W semaphore slow path. Here a waiting process is
2401queued on the semaphore, by virtue of it having a piece of its stack linked to
2402the semaphore's list of waiting processes:
2403
2404 struct rw_semaphore {
2405 ...
2406 spinlock_t lock;
2407 struct list_head waiters;
2408 };
2409
2410 struct rwsem_waiter {
2411 struct list_head list;
2412 struct task_struct *task;
2413 };
2414
2415To wake up a particular waiter, the up_read() or up_write() functions have to:
2416
2417 (1) read the next pointer from this waiter's record to know as to where the
2418 next waiter record is;
2419
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002420 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002421
2422 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2423
2424 (4) call wake_up_process() on the task; and
2425
2426 (5) release the reference held on the waiter's task struct.
2427
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002428In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002429
2430 LOAD waiter->list.next;
2431 LOAD waiter->task;
2432 STORE waiter->task;
2433 CALL wakeup
2434 RELEASE task
2435
2436and if any of these steps occur out of order, then the whole thing may
2437malfunction.
2438
2439Once it has queued itself and dropped the semaphore lock, the waiter does not
2440get the lock again; it instead just waits for its task pointer to be cleared
2441before proceeding. Since the record is on the waiter's stack, this means that
2442if the task pointer is cleared _before_ the next pointer in the list is read,
2443another CPU might start processing the waiter and might clobber the waiter's
2444stack before the up*() function has a chance to read the next pointer.
2445
2446Consider then what might happen to the above sequence of events:
2447
2448 CPU 1 CPU 2
2449 =============================== ===============================
2450 down_xxx()
2451 Queue waiter
2452 Sleep
2453 up_yyy()
2454 LOAD waiter->task;
2455 STORE waiter->task;
2456 Woken up by other event
2457 <preempt>
2458 Resume processing
2459 down_xxx() returns
2460 call foo()
2461 foo() clobbers *waiter
2462 </preempt>
2463 LOAD waiter->list.next;
2464 --- OOPS ---
2465
2466This could be dealt with using the semaphore lock, but then the down_xxx()
2467function has to needlessly get the spinlock again after being woken up.
2468
2469The way to deal with this is to insert a general SMP memory barrier:
2470
2471 LOAD waiter->list.next;
2472 LOAD waiter->task;
2473 smp_mb();
2474 STORE waiter->task;
2475 CALL wakeup
2476 RELEASE task
2477
2478In this case, the barrier makes a guarantee that all memory accesses before the
2479barrier will appear to happen before all the memory accesses after the barrier
2480with respect to the other CPUs on the system. It does _not_ guarantee that all
2481the memory accesses before the barrier will be complete by the time the barrier
2482instruction itself is complete.
2483
2484On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2485compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002486right order without actually intervening in the CPU. Since there's only one
2487CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002488
2489
2490ATOMIC OPERATIONS
2491-----------------
2492
David Howellsdbc87002006-04-10 22:54:23 -07002493Whilst they are technically interprocessor interaction considerations, atomic
2494operations are noted specially as some of them imply full memory barriers and
2495some don't, but they're very heavily relied on as a group throughout the
2496kernel.
2497
Peter Zijlstra706eeb32017-06-12 14:50:27 +02002498See Documentation/atomic_t.txt for more information.
David Howells108b42b2006-03-31 16:00:29 +01002499
2500
2501ACCESSING DEVICES
2502-----------------
2503
2504Many devices can be memory mapped, and so appear to the CPU as if they're just
2505a set of memory locations. To control such a device, the driver usually has to
2506make the right memory accesses in exactly the right order.
2507
2508However, having a clever CPU or a clever compiler creates a potential problem
2509in that the carefully sequenced accesses in the driver code won't reach the
2510device in the requisite order if the CPU or the compiler thinks it is more
2511efficient to reorder, combine or merge accesses - something that would cause
2512the device to malfunction.
2513
2514Inside of the Linux kernel, I/O should be done through the appropriate accessor
2515routines - such as inb() or writel() - which know how to make such accesses
2516appropriately sequential. Whilst this, for the most part, renders the explicit
2517use of memory barriers unnecessary, there are a couple of situations where they
2518might be needed:
2519
2520 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2521 so for _all_ general drivers locks should be used and mmiowb() must be
2522 issued prior to unlocking the critical section.
2523
2524 (2) If the accessor functions are used to refer to an I/O memory window with
2525 relaxed memory access properties, then _mandatory_ memory barriers are
2526 required to enforce ordering.
2527
Helmut Grohne0fe397f2017-05-03 11:51:46 +02002528See Documentation/driver-api/device-io.rst for more information.
David Howells108b42b2006-03-31 16:00:29 +01002529
2530
2531INTERRUPTS
2532----------
2533
2534A driver may be interrupted by its own interrupt service routine, and thus the
2535two parts of the driver may interfere with each other's attempts to control or
2536access the device.
2537
2538This may be alleviated - at least in part - by disabling local interrupts (a
2539form of locking), such that the critical operations are all contained within
2540the interrupt-disabled section in the driver. Whilst the driver's interrupt
2541routine is executing, the driver's core may not run on the same CPU, and its
2542interrupt is not permitted to happen again until the current interrupt has been
2543handled, thus the interrupt handler does not need to lock against that.
2544
2545However, consider a driver that was talking to an ethernet card that sports an
2546address register and a data register. If that driver's core talks to the card
2547under interrupt-disablement and then the driver's interrupt handler is invoked:
2548
2549 LOCAL IRQ DISABLE
2550 writew(ADDR, 3);
2551 writew(DATA, y);
2552 LOCAL IRQ ENABLE
2553 <interrupt>
2554 writew(ADDR, 4);
2555 q = readw(DATA);
2556 </interrupt>
2557
2558The store to the data register might happen after the second store to the
2559address register if ordering rules are sufficiently relaxed:
2560
2561 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2562
2563
2564If ordering rules are relaxed, it must be assumed that accesses done inside an
2565interrupt disabled section may leak outside of it and may interleave with
2566accesses performed in an interrupt - and vice versa - unless implicit or
2567explicit barriers are used.
2568
2569Normally this won't be a problem because the I/O accesses done inside such
2570sections will include synchronous load operations on strictly ordered I/O
SeongJae Park0b6fa342016-04-12 08:52:53 -07002571registers that form implicit I/O barriers. If this isn't sufficient then an
David Howells108b42b2006-03-31 16:00:29 +01002572mmiowb() may need to be used explicitly.
2573
2574
2575A similar situation may occur between an interrupt routine and two routines
SeongJae Park0b6fa342016-04-12 08:52:53 -07002576running on separate CPUs that communicate with each other. If such a case is
David Howells108b42b2006-03-31 16:00:29 +01002577likely, then interrupt-disabling locks should be used to guarantee ordering.
2578
2579
2580==========================
2581KERNEL I/O BARRIER EFFECTS
2582==========================
2583
2584When accessing I/O memory, drivers should use the appropriate accessor
2585functions:
2586
2587 (*) inX(), outX():
2588
2589 These are intended to talk to I/O space rather than memory space, but
SeongJae Park0b6fa342016-04-12 08:52:53 -07002590 that's primarily a CPU-specific concept. The i386 and x86_64 processors
2591 do indeed have special I/O space access cycles and instructions, but many
David Howells108b42b2006-03-31 16:00:29 +01002592 CPUs don't have such a concept.
2593
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002594 The PCI bus, amongst others, defines an I/O space concept which - on such
2595 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002596 space. However, it may also be mapped as a virtual I/O space in the CPU's
2597 memory map, particularly on those CPUs that don't support alternate I/O
2598 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002599
2600 Accesses to this space may be fully synchronous (as on i386), but
2601 intermediary bridges (such as the PCI host bridge) may not fully honour
2602 that.
2603
2604 They are guaranteed to be fully ordered with respect to each other.
2605
2606 They are not guaranteed to be fully ordered with respect to other types of
2607 memory and I/O operation.
2608
2609 (*) readX(), writeX():
2610
2611 Whether these are guaranteed to be fully ordered and uncombined with
2612 respect to each other on the issuing CPU depends on the characteristics
SeongJae Park0b6fa342016-04-12 08:52:53 -07002613 defined for the memory window through which they're accessing. On later
David Howells108b42b2006-03-31 16:00:29 +01002614 i386 architecture machines, for example, this is controlled by way of the
2615 MTRR registers.
2616
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002617 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002618 provided they're not accessing a prefetchable device.
2619
2620 However, intermediary hardware (such as a PCI bridge) may indulge in
2621 deferral if it so wishes; to flush a store, a load from the same location
2622 is preferred[*], but a load from the same device or from configuration
2623 space should suffice for PCI.
2624
2625 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002626 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2627 example.
David Howells108b42b2006-03-31 16:00:29 +01002628
2629 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2630 force stores to be ordered.
2631
2632 Please refer to the PCI specification for more information on interactions
2633 between PCI transactions.
2634
Will Deacona8e0aea2013-09-04 12:30:08 +01002635 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002636
Will Deacona8e0aea2013-09-04 12:30:08 +01002637 These are similar to readX() and writeX(), but provide weaker memory
SeongJae Park0b6fa342016-04-12 08:52:53 -07002638 ordering guarantees. Specifically, they do not guarantee ordering with
Will Deacona8e0aea2013-09-04 12:30:08 +01002639 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
SeongJae Park0b6fa342016-04-12 08:52:53 -07002640 ordering with respect to LOCK or UNLOCK operations. If the latter is
2641 required, an mmiowb() barrier can be used. Note that relaxed accesses to
Will Deacona8e0aea2013-09-04 12:30:08 +01002642 the same peripheral are guaranteed to be ordered with respect to each
2643 other.
David Howells108b42b2006-03-31 16:00:29 +01002644
2645 (*) ioreadX(), iowriteX()
2646
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002647 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002648 doing, be it inX()/outX() or readX()/writeX().
2649
2650
2651========================================
2652ASSUMED MINIMUM EXECUTION ORDERING MODEL
2653========================================
2654
2655It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2656maintain the appearance of program causality with respect to itself. Some CPUs
2657(such as i386 or x86_64) are more constrained than others (such as powerpc or
2658frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2659of arch-specific code.
2660
2661This means that it must be considered that the CPU will execute its instruction
2662stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002663instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002664earlier instruction must be sufficiently complete[*] before the later
2665instruction may proceed; in other words: provided that the appearance of
2666causality is maintained.
2667
2668 [*] Some instructions have more than one effect - such as changing the
2669 condition codes, changing registers or changing memory - and different
2670 instructions may depend on different effects.
2671
2672A CPU may also discard any instruction sequence that winds up having no
2673ultimate effect. For example, if two adjacent instructions both load an
2674immediate value into the same register, the first may be discarded.
2675
2676
2677Similarly, it has to be assumed that compiler might reorder the instruction
2678stream in any way it sees fit, again provided the appearance of causality is
2679maintained.
2680
2681
2682============================
2683THE EFFECTS OF THE CPU CACHE
2684============================
2685
2686The way cached memory operations are perceived across the system is affected to
2687a certain extent by the caches that lie between CPUs and memory, and by the
2688memory coherence system that maintains the consistency of state in the system.
2689
2690As far as the way a CPU interacts with another part of the system through the
2691caches goes, the memory system has to include the CPU's caches, and memory
2692barriers for the most part act at the interface between the CPU and its cache
2693(memory barriers logically act on the dotted line in the following diagram):
2694
2695 <--- CPU ---> : <----------- Memory ----------->
2696 :
2697 +--------+ +--------+ : +--------+ +-----------+
2698 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002699 | CPU | | Memory | : | CPU | | | | |
2700 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002701 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002702 | | | | : | | | | | |
2703 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002704 : | Cache | +--------+
2705 : | Coherency |
2706 : | Mechanism | +--------+
2707 +--------+ +--------+ : +--------+ | | | |
2708 | | | | : | | | | | |
2709 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002710 | Core |--->| Access |----->| Cache |<-->| | | |
2711 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002712 | | | | : | | | | +--------+
2713 +--------+ +--------+ : +--------+ +-----------+
2714 :
2715 :
2716
2717Although any particular load or store may not actually appear outside of the
2718CPU that issued it since it may have been satisfied within the CPU's own cache,
2719it will still appear as if the full memory access had taken place as far as the
2720other CPUs are concerned since the cache coherency mechanisms will migrate the
2721cacheline over to the accessing CPU and propagate the effects upon conflict.
2722
2723The CPU core may execute instructions in any order it deems fit, provided the
2724expected program causality appears to be maintained. Some of the instructions
2725generate load and store operations which then go into the queue of memory
2726accesses to be performed. The core may place these in the queue in any order
2727it wishes, and continue execution until it is forced to wait for an instruction
2728to complete.
2729
2730What memory barriers are concerned with is controlling the order in which
2731accesses cross from the CPU side of things to the memory side of things, and
2732the order in which the effects are perceived to happen by the other observers
2733in the system.
2734
2735[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2736their own loads and stores as if they had happened in program order.
2737
2738[!] MMIO or other device accesses may bypass the cache system. This depends on
2739the properties of the memory window through which devices are accessed and/or
2740the use of any special device communication instructions the CPU may have.
2741
2742
2743CACHE COHERENCY
2744---------------
2745
2746Life isn't quite as simple as it may appear above, however: for while the
2747caches are expected to be coherent, there's no guarantee that that coherency
2748will be ordered. This means that whilst changes made on one CPU will
2749eventually become visible on all CPUs, there's no guarantee that they will
2750become apparent in the same order on those other CPUs.
2751
2752
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002753Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2754has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002755
2756 :
2757 : +--------+
2758 : +---------+ | |
2759 +--------+ : +--->| Cache A |<------->| |
2760 | | : | +---------+ | |
2761 | CPU 1 |<---+ | |
2762 | | : | +---------+ | |
2763 +--------+ : +--->| Cache B |<------->| |
2764 : +---------+ | |
2765 : | Memory |
2766 : +---------+ | System |
2767 +--------+ : +--->| Cache C |<------->| |
2768 | | : | +---------+ | |
2769 | CPU 2 |<---+ | |
2770 | | : | +---------+ | |
2771 +--------+ : +--->| Cache D |<------->| |
2772 : +---------+ | |
2773 : +--------+
2774 :
2775
2776Imagine the system has the following properties:
2777
2778 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2779 resident in memory;
2780
2781 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2782 resident in memory;
2783
2784 (*) whilst the CPU core is interrogating one cache, the other cache may be
2785 making use of the bus to access the rest of the system - perhaps to
2786 displace a dirty cacheline or to do a speculative load;
2787
2788 (*) each cache has a queue of operations that need to be applied to that cache
2789 to maintain coherency with the rest of the system;
2790
2791 (*) the coherency queue is not flushed by normal loads to lines already
2792 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002793 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002794
2795Imagine, then, that two writes are made on the first CPU, with a write barrier
2796between them to guarantee that they will appear to reach that CPU's caches in
2797the requisite order:
2798
2799 CPU 1 CPU 2 COMMENT
2800 =============== =============== =======================================
2801 u == 0, v == 1 and p == &u, q == &u
2802 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002803 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002804 change to p
2805 <A:modify v=2> v is now in cache A exclusively
2806 p = &v;
2807 <B:modify p=&v> p is now in cache B exclusively
2808
2809The write memory barrier forces the other CPUs in the system to perceive that
2810the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002811now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002812
2813 CPU 1 CPU 2 COMMENT
2814 =============== =============== =======================================
2815 ...
2816 q = p;
2817 x = *q;
2818
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002819The above pair of reads may then fail to happen in the expected order, as the
David Howells108b42b2006-03-31 16:00:29 +01002820cacheline holding p may get updated in one of the second CPU's caches whilst
2821the update to the cacheline holding v is delayed in the other of the second
2822CPU's caches by some other cache event:
2823
2824 CPU 1 CPU 2 COMMENT
2825 =============== =============== =======================================
2826 u == 0, v == 1 and p == &u, q == &u
2827 v = 2;
2828 smp_wmb();
2829 <A:modify v=2> <C:busy>
2830 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002831 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002832 <D:request p>
2833 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002834 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002835 x = *q;
2836 <C:read *q> Reads from v before v updated in cache
2837 <C:unbusy>
2838 <C:commit v=2>
2839
2840Basically, whilst both cachelines will be updated on CPU 2 eventually, there's
2841no guarantee that, without intervention, the order of update will be the same
2842as that committed on CPU 1.
2843
2844
2845To intervene, we need to interpolate a data dependency barrier or a read
2846barrier between the loads. This will force the cache to commit its coherency
2847queue before processing any further requests:
2848
2849 CPU 1 CPU 2 COMMENT
2850 =============== =============== =======================================
2851 u == 0, v == 1 and p == &u, q == &u
2852 v = 2;
2853 smp_wmb();
2854 <A:modify v=2> <C:busy>
2855 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002856 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002857 <D:request p>
2858 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002859 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002860 smp_read_barrier_depends()
2861 <C:unbusy>
2862 <C:commit v=2>
2863 x = *q;
2864 <C:read *q> Reads from v after v updated in cache
2865
2866
2867This sort of problem can be encountered on DEC Alpha processors as they have a
2868split cache that improves performance by making better use of the data bus.
2869Whilst most CPUs do imply a data dependency barrier on the read when a memory
2870access depends on a read, not all do, so it may not be relied on.
2871
2872Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002873cachelets for normal memory accesses. The semantics of the Alpha removes the
Paul E. McKenney9ad3c142017-11-27 09:20:40 -08002874need for hardware coordination in the absence of memory barriers, which
2875permitted Alpha to sport higher CPU clock rates back in the day. However,
2876please note that smp_read_barrier_depends() should not be used except in
2877Alpha arch-specific code and within the READ_ONCE() macro.
David Howells108b42b2006-03-31 16:00:29 +01002878
2879
2880CACHE COHERENCY VS DMA
2881----------------------
2882
2883Not all systems maintain cache coherency with respect to devices doing DMA. In
2884such cases, a device attempting DMA may obtain stale data from RAM because
2885dirty cache lines may be resident in the caches of various CPUs, and may not
2886have been written back to RAM yet. To deal with this, the appropriate part of
2887the kernel must flush the overlapping bits of cache on each CPU (and maybe
2888invalidate them as well).
2889
2890In addition, the data DMA'd to RAM by a device may be overwritten by dirty
2891cache lines being written back to RAM from a CPU's cache after the device has
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002892installed its own data, or cache lines present in the CPU's cache may simply
2893obscure the fact that RAM has been updated, until at such time as the cacheline
2894is discarded from the CPU's cache and reloaded. To deal with this, the
2895appropriate part of the kernel must invalidate the overlapping bits of the
David Howells108b42b2006-03-31 16:00:29 +01002896cache on each CPU.
2897
2898See Documentation/cachetlb.txt for more information on cache management.
2899
2900
2901CACHE COHERENCY VS MMIO
2902-----------------------
2903
2904Memory mapped I/O usually takes place through memory locations that are part of
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002905a window in the CPU's memory space that has different properties assigned than
David Howells108b42b2006-03-31 16:00:29 +01002906the usual RAM directed window.
2907
2908Amongst these properties is usually the fact that such accesses bypass the
2909caching entirely and go directly to the device buses. This means MMIO accesses
2910may, in effect, overtake accesses to cached memory that were emitted earlier.
2911A memory barrier isn't sufficient in such a case, but rather the cache must be
2912flushed between the cached memory write and the MMIO access if the two are in
2913any way dependent.
2914
2915
2916=========================
2917THE THINGS CPUS GET UP TO
2918=========================
2919
2920A programmer might take it for granted that the CPU will perform memory
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002921operations in exactly the order specified, so that if the CPU is, for example,
David Howells108b42b2006-03-31 16:00:29 +01002922given the following piece of code to execute:
2923
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002924 a = READ_ONCE(*A);
2925 WRITE_ONCE(*B, b);
2926 c = READ_ONCE(*C);
2927 d = READ_ONCE(*D);
2928 WRITE_ONCE(*E, e);
David Howells108b42b2006-03-31 16:00:29 +01002929
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002930they would then expect that the CPU will complete the memory operation for each
David Howells108b42b2006-03-31 16:00:29 +01002931instruction before moving on to the next one, leading to a definite sequence of
2932operations as seen by external observers in the system:
2933
2934 LOAD *A, STORE *B, LOAD *C, LOAD *D, STORE *E.
2935
2936
2937Reality is, of course, much messier. With many CPUs and compilers, the above
2938assumption doesn't hold because:
2939
2940 (*) loads are more likely to need to be completed immediately to permit
2941 execution progress, whereas stores can often be deferred without a
2942 problem;
2943
2944 (*) loads may be done speculatively, and the result discarded should it prove
2945 to have been unnecessary;
2946
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002947 (*) loads may be done speculatively, leading to the result having been fetched
2948 at the wrong time in the expected sequence of events;
David Howells108b42b2006-03-31 16:00:29 +01002949
2950 (*) the order of the memory accesses may be rearranged to promote better use
2951 of the CPU buses and caches;
2952
2953 (*) loads and stores may be combined to improve performance when talking to
2954 memory or I/O hardware that can do batched accesses of adjacent locations,
2955 thus cutting down on transaction setup costs (memory and PCI devices may
2956 both be able to do this); and
2957
2958 (*) the CPU's data cache may affect the ordering, and whilst cache-coherency
2959 mechanisms may alleviate this - once the store has actually hit the cache
2960 - there's no guarantee that the coherency management will be propagated in
2961 order to other CPUs.
2962
2963So what another CPU, say, might actually observe from the above piece of code
2964is:
2965
2966 LOAD *A, ..., LOAD {*C,*D}, STORE *E, STORE *B
2967
2968 (Where "LOAD {*C,*D}" is a combined load)
2969
2970
2971However, it is guaranteed that a CPU will be self-consistent: it will see its
2972_own_ accesses appear to be correctly ordered, without the need for a memory
2973barrier. For instance with the following code:
2974
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002975 U = READ_ONCE(*A);
2976 WRITE_ONCE(*A, V);
2977 WRITE_ONCE(*A, W);
2978 X = READ_ONCE(*A);
2979 WRITE_ONCE(*A, Y);
2980 Z = READ_ONCE(*A);
David Howells108b42b2006-03-31 16:00:29 +01002981
2982and assuming no intervention by an external influence, it can be assumed that
2983the final result will appear to be:
2984
2985 U == the original value of *A
2986 X == W
2987 Z == Y
2988 *A == Y
2989
2990The code above may cause the CPU to generate the full sequence of memory
2991accesses:
2992
2993 U=LOAD *A, STORE *A=V, STORE *A=W, X=LOAD *A, STORE *A=Y, Z=LOAD *A
2994
2995in that order, but, without intervention, the sequence may have almost any
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002996combination of elements combined or discarded, provided the program's view
2997of the world remains consistent. Note that READ_ONCE() and WRITE_ONCE()
2998are -not- optional in the above example, as there are architectures
2999where a given CPU might reorder successive loads to the same location.
3000On such architectures, READ_ONCE() and WRITE_ONCE() do whatever is
3001necessary to prevent this, for example, on Itanium the volatile casts
3002used by READ_ONCE() and WRITE_ONCE() cause GCC to emit the special ld.acq
3003and st.rel instructions (respectively) that prevent such reordering.
David Howells108b42b2006-03-31 16:00:29 +01003004
3005The compiler may also combine, discard or defer elements of the sequence before
3006the CPU even sees them.
3007
3008For instance:
3009
3010 *A = V;
3011 *A = W;
3012
3013may be reduced to:
3014
3015 *A = W;
3016
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003017since, without either a write barrier or an WRITE_ONCE(), it can be
Paul E. McKenney2ecf8102013-12-11 13:59:04 -08003018assumed that the effect of the storage of V to *A is lost. Similarly:
David Howells108b42b2006-03-31 16:00:29 +01003019
3020 *A = Y;
3021 Z = *A;
3022
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003023may, without a memory barrier or an READ_ONCE() and WRITE_ONCE(), be
3024reduced to:
David Howells108b42b2006-03-31 16:00:29 +01003025
3026 *A = Y;
3027 Z = Y;
3028
3029and the LOAD operation never appear outside of the CPU.
3030
3031
3032AND THEN THERE'S THE ALPHA
3033--------------------------
3034
3035The DEC Alpha CPU is one of the most relaxed CPUs there is. Not only that,
3036some versions of the Alpha CPU have a split data cache, permitting them to have
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003037two semantically-related cache lines updated at separate times. This is where
David Howells108b42b2006-03-31 16:00:29 +01003038the data dependency barrier really becomes necessary as this synchronises both
3039caches with the memory coherence system, thus making it seem like pointer
3040changes vs new data occur in the right order.
3041
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003042The Alpha defines the Linux kernel's memory barrier model.
David Howells108b42b2006-03-31 16:00:29 +01003043
3044See the subsection on "Cache Coherency" above.
3045
SeongJae Park0b6fa342016-04-12 08:52:53 -07003046
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003047VIRTUAL MACHINE GUESTS
SeongJae Park3dbf0912016-04-12 08:52:52 -07003048----------------------
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003049
3050Guests running within virtual machines might be affected by SMP effects even if
3051the guest itself is compiled without SMP support. This is an artifact of
3052interfacing with an SMP host while running an UP kernel. Using mandatory
3053barriers for this use-case would be possible but is often suboptimal.
3054
3055To handle this case optimally, low-level virt_mb() etc macros are available.
3056These have the same effect as smp_mb() etc when SMP is enabled, but generate
SeongJae Park0b6fa342016-04-12 08:52:53 -07003057identical code for SMP and non-SMP systems. For example, virtual machine guests
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003058should use virt_mb() rather than smp_mb() when synchronizing against a
3059(possibly SMP) host.
3060
3061These are equivalent to smp_mb() etc counterparts in all other respects,
3062in particular, they do not control MMIO effects: to control
3063MMIO effects, use mandatory barriers.
David Howells108b42b2006-03-31 16:00:29 +01003064
SeongJae Park0b6fa342016-04-12 08:52:53 -07003065
David Howells90fddab2010-03-24 09:43:00 +00003066============
3067EXAMPLE USES
3068============
3069
3070CIRCULAR BUFFERS
3071----------------
3072
3073Memory barriers can be used to implement circular buffering without the need
3074of a lock to serialise the producer with the consumer. See:
3075
3076 Documentation/circular-buffers.txt
3077
3078for details.
3079
3080
David Howells108b42b2006-03-31 16:00:29 +01003081==========
3082REFERENCES
3083==========
3084
3085Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
3086Digital Press)
3087 Chapter 5.2: Physical Address Space Characteristics
3088 Chapter 5.4: Caches and Write Buffers
3089 Chapter 5.5: Data Sharing
3090 Chapter 5.6: Read/Write Ordering
3091
3092AMD64 Architecture Programmer's Manual Volume 2: System Programming
3093 Chapter 7.1: Memory-Access Ordering
3094 Chapter 7.4: Buffering and Combining Memory Writes
3095
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07003096ARM Architecture Reference Manual (ARMv8, for ARMv8-A architecture profile)
3097 Chapter B2: The AArch64 Application Level Memory Model
3098
David Howells108b42b2006-03-31 16:00:29 +01003099IA-32 Intel Architecture Software Developer's Manual, Volume 3:
3100System Programming Guide
3101 Chapter 7.1: Locked Atomic Operations
3102 Chapter 7.2: Memory Ordering
3103 Chapter 7.4: Serializing Instructions
3104
3105The SPARC Architecture Manual, Version 9
3106 Chapter 8: Memory Models
3107 Appendix D: Formal Specification of the Memory Models
3108 Appendix J: Programming with the Memory Models
3109
Paul E. McKenneyf1ab25a2017-08-29 15:49:21 -07003110Storage in the PowerPC (Stone and Fitzgerald)
3111
David Howells108b42b2006-03-31 16:00:29 +01003112UltraSPARC Programmer Reference Manual
3113 Chapter 5: Memory Accesses and Cacheability
3114 Chapter 15: Sparc-V9 Memory Models
3115
3116UltraSPARC III Cu User's Manual
3117 Chapter 9: Memory Models
3118
3119UltraSPARC IIIi Processor User's Manual
3120 Chapter 8: Memory Models
3121
3122UltraSPARC Architecture 2005
3123 Chapter 9: Memory
3124 Appendix D: Formal Specifications of the Memory Models
3125
3126UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3127 Chapter 8: Memory Models
3128 Appendix F: Caches and Cache Coherency
3129
3130Solaris Internals, Core Kernel Architecture, p63-68:
3131 Chapter 3.3: Hardware Considerations for Locks and
3132 Synchronization
3133
3134Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3135for Kernel Programmers:
3136 Chapter 13: Other Memory Models
3137
3138Intel Itanium Architecture Software Developer's Manual: Volume 1:
3139 Section 2.6: Speculation
3140 Section 4.4: Memory Access