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David Howells108b42b2006-03-31 16:00:29 +01001 ============================
2 LINUX KERNEL MEMORY BARRIERS
3 ============================
4
5By: David Howells <dhowells@redhat.com>
David Howells90fddab2010-03-24 09:43:00 +00006 Paul E. McKenney <paulmck@linux.vnet.ibm.com>
David Howells108b42b2006-03-31 16:00:29 +01007
8Contents:
9
10 (*) Abstract memory access model.
11
12 - Device operations.
13 - Guarantees.
14
15 (*) What are memory barriers?
16
17 - Varieties of memory barrier.
18 - What may not be assumed about memory barriers?
19 - Data dependency barriers.
20 - Control dependencies.
21 - SMP barrier pairing.
22 - Examples of memory barrier sequences.
David Howells670bd952006-06-10 09:54:12 -070023 - Read memory barriers vs load speculation.
Paul E. McKenney241e6662011-02-10 16:54:50 -080024 - Transitivity
David Howells108b42b2006-03-31 16:00:29 +010025
26 (*) Explicit kernel barriers.
27
28 - Compiler barrier.
Jarek Poplawski81fc6322007-05-23 13:58:20 -070029 - CPU memory barriers.
David Howells108b42b2006-03-31 16:00:29 +010030 - MMIO write barrier.
31
32 (*) Implicit kernel memory barriers.
33
SeongJae Park166bda72016-04-12 08:52:50 -070034 - Lock acquisition functions.
David Howells108b42b2006-03-31 16:00:29 +010035 - Interrupt disabling functions.
David Howells50fa6102009-04-28 15:01:38 +010036 - Sleep and wake-up functions.
David Howells108b42b2006-03-31 16:00:29 +010037 - Miscellaneous functions.
38
SeongJae Park166bda72016-04-12 08:52:50 -070039 (*) Inter-CPU acquiring barrier effects.
David Howells108b42b2006-03-31 16:00:29 +010040
SeongJae Park166bda72016-04-12 08:52:50 -070041 - Acquires vs memory accesses.
42 - Acquires vs I/O accesses.
David Howells108b42b2006-03-31 16:00:29 +010043
44 (*) Where are memory barriers needed?
45
46 - Interprocessor interaction.
47 - Atomic operations.
48 - Accessing devices.
49 - Interrupts.
50
51 (*) Kernel I/O barrier effects.
52
53 (*) Assumed minimum execution ordering model.
54
55 (*) The effects of the cpu cache.
56
57 - Cache coherency.
58 - Cache coherency vs DMA.
59 - Cache coherency vs MMIO.
60
61 (*) The things CPUs get up to.
62
63 - And then there's the Alpha.
SeongJae Park01e1cd62016-04-12 08:52:51 -070064 - Virtual Machine Guests.
David Howells108b42b2006-03-31 16:00:29 +010065
David Howells90fddab2010-03-24 09:43:00 +000066 (*) Example uses.
67
68 - Circular buffers.
69
David Howells108b42b2006-03-31 16:00:29 +010070 (*) References.
71
72
73============================
74ABSTRACT MEMORY ACCESS MODEL
75============================
76
77Consider the following abstract model of the system:
78
79 : :
80 : :
81 : :
82 +-------+ : +--------+ : +-------+
83 | | : | | : | |
84 | | : | | : | |
85 | CPU 1 |<----->| Memory |<----->| CPU 2 |
86 | | : | | : | |
87 | | : | | : | |
88 +-------+ : +--------+ : +-------+
89 ^ : ^ : ^
90 | : | : |
91 | : | : |
92 | : v : |
93 | : +--------+ : |
94 | : | | : |
95 | : | | : |
96 +---------->| Device |<----------+
97 : | | :
98 : | | :
99 : +--------+ :
100 : :
101
102Each CPU executes a program that generates memory access operations. In the
103abstract CPU, memory operation ordering is very relaxed, and a CPU may actually
104perform the memory operations in any order it likes, provided program causality
105appears to be maintained. Similarly, the compiler may also arrange the
106instructions it emits in any order it likes, provided it doesn't affect the
107apparent operation of the program.
108
109So in the above diagram, the effects of the memory operations performed by a
110CPU are perceived by the rest of the system as the operations cross the
111interface between the CPU and rest of the system (the dotted lines).
112
113
114For example, consider the following sequence of events:
115
116 CPU 1 CPU 2
117 =============== ===============
118 { A == 1; B == 2 }
Alexey Dobriyan615cc2c2014-06-06 14:36:41 -0700119 A = 3; x = B;
120 B = 4; y = A;
David Howells108b42b2006-03-31 16:00:29 +0100121
122The set of accesses as seen by the memory system in the middle can be arranged
123in 24 different combinations:
124
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400125 STORE A=3, STORE B=4, y=LOAD A->3, x=LOAD B->4
126 STORE A=3, STORE B=4, x=LOAD B->4, y=LOAD A->3
127 STORE A=3, y=LOAD A->3, STORE B=4, x=LOAD B->4
128 STORE A=3, y=LOAD A->3, x=LOAD B->2, STORE B=4
129 STORE A=3, x=LOAD B->2, STORE B=4, y=LOAD A->3
130 STORE A=3, x=LOAD B->2, y=LOAD A->3, STORE B=4
131 STORE B=4, STORE A=3, y=LOAD A->3, x=LOAD B->4
David Howells108b42b2006-03-31 16:00:29 +0100132 STORE B=4, ...
133 ...
134
135and can thus result in four different combinations of values:
136
Pranith Kumar8ab8b3e2014-09-02 23:34:29 -0400137 x == 2, y == 1
138 x == 2, y == 3
139 x == 4, y == 1
140 x == 4, y == 3
David Howells108b42b2006-03-31 16:00:29 +0100141
142
143Furthermore, the stores committed by a CPU to the memory system may not be
144perceived by the loads made by another CPU in the same order as the stores were
145committed.
146
147
148As a further example, consider this sequence of events:
149
150 CPU 1 CPU 2
151 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700152 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100153 B = 4; Q = P;
154 P = &B D = *Q;
155
156There is an obvious data dependency here, as the value loaded into D depends on
157the address retrieved from P by CPU 2. At the end of the sequence, any of the
158following results are possible:
159
160 (Q == &A) and (D == 1)
161 (Q == &B) and (D == 2)
162 (Q == &B) and (D == 4)
163
164Note that CPU 2 will never try and load C into D because the CPU will load P
165into Q before issuing the load of *Q.
166
167
168DEVICE OPERATIONS
169-----------------
170
171Some devices present their control interfaces as collections of memory
172locations, but the order in which the control registers are accessed is very
173important. For instance, imagine an ethernet card with a set of internal
174registers that are accessed through an address port register (A) and a data
175port register (D). To read internal register 5, the following code might then
176be used:
177
178 *A = 5;
179 x = *D;
180
181but this might show up as either of the following two sequences:
182
183 STORE *A = 5, x = LOAD *D
184 x = LOAD *D, STORE *A = 5
185
186the second of which will almost certainly result in a malfunction, since it set
187the address _after_ attempting to read the register.
188
189
190GUARANTEES
191----------
192
193There are some minimal guarantees that may be expected of a CPU:
194
195 (*) On any given CPU, dependent memory accesses will be issued in order, with
196 respect to itself. This means that for:
197
Chris Metcalff84cfbb2015-11-23 17:04:17 -0500198 Q = READ_ONCE(P); smp_read_barrier_depends(); D = READ_ONCE(*Q);
David Howells108b42b2006-03-31 16:00:29 +0100199
200 the CPU will issue the following memory operations:
201
202 Q = LOAD P, D = LOAD *Q
203
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800204 and always in that order. On most systems, smp_read_barrier_depends()
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700205 does nothing, but it is required for DEC Alpha. The READ_ONCE()
Chris Metcalff84cfbb2015-11-23 17:04:17 -0500206 is required to prevent compiler mischief. Please note that you
207 should normally use something like rcu_dereference() instead of
208 open-coding smp_read_barrier_depends().
David Howells108b42b2006-03-31 16:00:29 +0100209
210 (*) Overlapping loads and stores within a particular CPU will appear to be
211 ordered within that CPU. This means that for:
212
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700213 a = READ_ONCE(*X); WRITE_ONCE(*X, b);
David Howells108b42b2006-03-31 16:00:29 +0100214
215 the CPU will only issue the following sequence of memory operations:
216
217 a = LOAD *X, STORE *X = b
218
219 And for:
220
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700221 WRITE_ONCE(*X, c); d = READ_ONCE(*X);
David Howells108b42b2006-03-31 16:00:29 +0100222
223 the CPU will only issue:
224
225 STORE *X = c, d = LOAD *X
226
Matt LaPlantefa00e7e2006-11-30 04:55:36 +0100227 (Loads and stores overlap if they are targeted at overlapping pieces of
David Howells108b42b2006-03-31 16:00:29 +0100228 memory).
229
230And there are a number of things that _must_ or _must_not_ be assumed:
231
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700232 (*) It _must_not_ be assumed that the compiler will do what you want
233 with memory references that are not protected by READ_ONCE() and
234 WRITE_ONCE(). Without them, the compiler is within its rights to
235 do all sorts of "creative" transformations, which are covered in
Paul E. McKenney895f5542016-01-06 14:23:03 -0800236 the COMPILER BARRIER section.
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800237
David Howells108b42b2006-03-31 16:00:29 +0100238 (*) It _must_not_ be assumed that independent loads and stores will be issued
239 in the order given. This means that for:
240
241 X = *A; Y = *B; *D = Z;
242
243 we may get any of the following sequences:
244
245 X = LOAD *A, Y = LOAD *B, STORE *D = Z
246 X = LOAD *A, STORE *D = Z, Y = LOAD *B
247 Y = LOAD *B, X = LOAD *A, STORE *D = Z
248 Y = LOAD *B, STORE *D = Z, X = LOAD *A
249 STORE *D = Z, X = LOAD *A, Y = LOAD *B
250 STORE *D = Z, Y = LOAD *B, X = LOAD *A
251
252 (*) It _must_ be assumed that overlapping memory accesses may be merged or
253 discarded. This means that for:
254
255 X = *A; Y = *(A + 4);
256
257 we may get any one of the following sequences:
258
259 X = LOAD *A; Y = LOAD *(A + 4);
260 Y = LOAD *(A + 4); X = LOAD *A;
261 {X, Y} = LOAD {*A, *(A + 4) };
262
263 And for:
264
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700265 *A = X; *(A + 4) = Y;
David Howells108b42b2006-03-31 16:00:29 +0100266
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700267 we may get any of:
David Howells108b42b2006-03-31 16:00:29 +0100268
Paul E. McKenneyf191eec2012-10-03 10:28:30 -0700269 STORE *A = X; STORE *(A + 4) = Y;
270 STORE *(A + 4) = Y; STORE *A = X;
271 STORE {*A, *(A + 4) } = {X, Y};
David Howells108b42b2006-03-31 16:00:29 +0100272
Paul E. McKenney432fbf32014-09-04 17:12:49 -0700273And there are anti-guarantees:
274
275 (*) These guarantees do not apply to bitfields, because compilers often
276 generate code to modify these using non-atomic read-modify-write
277 sequences. Do not attempt to use bitfields to synchronize parallel
278 algorithms.
279
280 (*) Even in cases where bitfields are protected by locks, all fields
281 in a given bitfield must be protected by one lock. If two fields
282 in a given bitfield are protected by different locks, the compiler's
283 non-atomic read-modify-write sequences can cause an update to one
284 field to corrupt the value of an adjacent field.
285
286 (*) These guarantees apply only to properly aligned and sized scalar
287 variables. "Properly sized" currently means variables that are
288 the same size as "char", "short", "int" and "long". "Properly
289 aligned" means the natural alignment, thus no constraints for
290 "char", two-byte alignment for "short", four-byte alignment for
291 "int", and either four-byte or eight-byte alignment for "long",
292 on 32-bit and 64-bit systems, respectively. Note that these
293 guarantees were introduced into the C11 standard, so beware when
294 using older pre-C11 compilers (for example, gcc 4.6). The portion
295 of the standard containing this guarantee is Section 3.14, which
296 defines "memory location" as follows:
297
298 memory location
299 either an object of scalar type, or a maximal sequence
300 of adjacent bit-fields all having nonzero width
301
302 NOTE 1: Two threads of execution can update and access
303 separate memory locations without interfering with
304 each other.
305
306 NOTE 2: A bit-field and an adjacent non-bit-field member
307 are in separate memory locations. The same applies
308 to two bit-fields, if one is declared inside a nested
309 structure declaration and the other is not, or if the two
310 are separated by a zero-length bit-field declaration,
311 or if they are separated by a non-bit-field member
312 declaration. It is not safe to concurrently update two
313 bit-fields in the same structure if all members declared
314 between them are also bit-fields, no matter what the
315 sizes of those intervening bit-fields happen to be.
316
David Howells108b42b2006-03-31 16:00:29 +0100317
318=========================
319WHAT ARE MEMORY BARRIERS?
320=========================
321
322As can be seen above, independent memory operations are effectively performed
323in random order, but this can be a problem for CPU-CPU interaction and for I/O.
324What is required is some way of intervening to instruct the compiler and the
325CPU to restrict the order.
326
327Memory barriers are such interventions. They impose a perceived partial
David Howells2b948952006-06-25 05:48:49 -0700328ordering over the memory operations on either side of the barrier.
329
330Such enforcement is important because the CPUs and other devices in a system
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700331can use a variety of tricks to improve performance, including reordering,
David Howells2b948952006-06-25 05:48:49 -0700332deferral and combination of memory operations; speculative loads; speculative
333branch prediction and various types of caching. Memory barriers are used to
334override or suppress these tricks, allowing the code to sanely control the
335interaction of multiple CPUs and/or devices.
David Howells108b42b2006-03-31 16:00:29 +0100336
337
338VARIETIES OF MEMORY BARRIER
339---------------------------
340
341Memory barriers come in four basic varieties:
342
343 (1) Write (or store) memory barriers.
344
345 A write memory barrier gives a guarantee that all the STORE operations
346 specified before the barrier will appear to happen before all the STORE
347 operations specified after the barrier with respect to the other
348 components of the system.
349
350 A write barrier is a partial ordering on stores only; it is not required
351 to have any effect on loads.
352
David Howells6bc39272006-06-25 05:49:22 -0700353 A CPU can be viewed as committing a sequence of store operations to the
David Howells108b42b2006-03-31 16:00:29 +0100354 memory system as time progresses. All stores before a write barrier will
355 occur in the sequence _before_ all the stores after the write barrier.
356
357 [!] Note that write barriers should normally be paired with read or data
358 dependency barriers; see the "SMP barrier pairing" subsection.
359
360
361 (2) Data dependency barriers.
362
363 A data dependency barrier is a weaker form of read barrier. In the case
364 where two loads are performed such that the second depends on the result
365 of the first (eg: the first load retrieves the address to which the second
366 load will be directed), a data dependency barrier would be required to
367 make sure that the target of the second load is updated before the address
368 obtained by the first load is accessed.
369
370 A data dependency barrier is a partial ordering on interdependent loads
371 only; it is not required to have any effect on stores, independent loads
372 or overlapping loads.
373
374 As mentioned in (1), the other CPUs in the system can be viewed as
375 committing sequences of stores to the memory system that the CPU being
376 considered can then perceive. A data dependency barrier issued by the CPU
377 under consideration guarantees that for any load preceding it, if that
378 load touches one of a sequence of stores from another CPU, then by the
379 time the barrier completes, the effects of all the stores prior to that
380 touched by the load will be perceptible to any loads issued after the data
381 dependency barrier.
382
383 See the "Examples of memory barrier sequences" subsection for diagrams
384 showing the ordering constraints.
385
386 [!] Note that the first load really has to have a _data_ dependency and
387 not a control dependency. If the address for the second load is dependent
388 on the first load, but the dependency is through a conditional rather than
389 actually loading the address itself, then it's a _control_ dependency and
390 a full read barrier or better is required. See the "Control dependencies"
391 subsection for more information.
392
393 [!] Note that data dependency barriers should normally be paired with
394 write barriers; see the "SMP barrier pairing" subsection.
395
396
397 (3) Read (or load) memory barriers.
398
399 A read barrier is a data dependency barrier plus a guarantee that all the
400 LOAD operations specified before the barrier will appear to happen before
401 all the LOAD operations specified after the barrier with respect to the
402 other components of the system.
403
404 A read barrier is a partial ordering on loads only; it is not required to
405 have any effect on stores.
406
407 Read memory barriers imply data dependency barriers, and so can substitute
408 for them.
409
410 [!] Note that read barriers should normally be paired with write barriers;
411 see the "SMP barrier pairing" subsection.
412
413
414 (4) General memory barriers.
415
David Howells670bd952006-06-10 09:54:12 -0700416 A general memory barrier gives a guarantee that all the LOAD and STORE
417 operations specified before the barrier will appear to happen before all
418 the LOAD and STORE operations specified after the barrier with respect to
419 the other components of the system.
420
421 A general memory barrier is a partial ordering over both loads and stores.
David Howells108b42b2006-03-31 16:00:29 +0100422
423 General memory barriers imply both read and write memory barriers, and so
424 can substitute for either.
425
426
427And a couple of implicit varieties:
428
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100429 (5) ACQUIRE operations.
David Howells108b42b2006-03-31 16:00:29 +0100430
431 This acts as a one-way permeable barrier. It guarantees that all memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100432 operations after the ACQUIRE operation will appear to happen after the
433 ACQUIRE operation with respect to the other components of the system.
Davidlohr Bueso787df632016-04-12 08:52:55 -0700434 ACQUIRE operations include LOCK operations and both smp_load_acquire()
435 and smp_cond_acquire() operations. The later builds the necessary ACQUIRE
436 semantics from relying on a control dependency and smp_rmb().
David Howells108b42b2006-03-31 16:00:29 +0100437
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100438 Memory operations that occur before an ACQUIRE operation may appear to
439 happen after it completes.
David Howells108b42b2006-03-31 16:00:29 +0100440
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100441 An ACQUIRE operation should almost always be paired with a RELEASE
442 operation.
David Howells108b42b2006-03-31 16:00:29 +0100443
444
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100445 (6) RELEASE operations.
David Howells108b42b2006-03-31 16:00:29 +0100446
447 This also acts as a one-way permeable barrier. It guarantees that all
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100448 memory operations before the RELEASE operation will appear to happen
449 before the RELEASE operation with respect to the other components of the
450 system. RELEASE operations include UNLOCK operations and
451 smp_store_release() operations.
David Howells108b42b2006-03-31 16:00:29 +0100452
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100453 Memory operations that occur after a RELEASE operation may appear to
David Howells108b42b2006-03-31 16:00:29 +0100454 happen before it completes.
455
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100456 The use of ACQUIRE and RELEASE operations generally precludes the need
457 for other sorts of memory barrier (but note the exceptions mentioned in
458 the subsection "MMIO write barrier"). In addition, a RELEASE+ACQUIRE
459 pair is -not- guaranteed to act as a full memory barrier. However, after
460 an ACQUIRE on a given variable, all memory accesses preceding any prior
461 RELEASE on that same variable are guaranteed to be visible. In other
462 words, within a given variable's critical section, all accesses of all
463 previous critical sections for that variable are guaranteed to have
464 completed.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -0800465
Peter Zijlstra2e4f5382013-11-06 14:57:36 +0100466 This means that ACQUIRE acts as a minimal "acquire" operation and
467 RELEASE acts as a minimal "release" operation.
David Howells108b42b2006-03-31 16:00:29 +0100468
469
470Memory barriers are only required where there's a possibility of interaction
471between two CPUs or between a CPU and a device. If it can be guaranteed that
472there won't be any such interaction in any particular piece of code, then
473memory barriers are unnecessary in that piece of code.
474
475
476Note that these are the _minimum_ guarantees. Different architectures may give
477more substantial guarantees, but they may _not_ be relied upon outside of arch
478specific code.
479
480
481WHAT MAY NOT BE ASSUMED ABOUT MEMORY BARRIERS?
482----------------------------------------------
483
484There are certain things that the Linux kernel memory barriers do not guarantee:
485
486 (*) There is no guarantee that any of the memory accesses specified before a
487 memory barrier will be _complete_ by the completion of a memory barrier
488 instruction; the barrier can be considered to draw a line in that CPU's
489 access queue that accesses of the appropriate type may not cross.
490
491 (*) There is no guarantee that issuing a memory barrier on one CPU will have
492 any direct effect on another CPU or any other hardware in the system. The
493 indirect effect will be the order in which the second CPU sees the effects
494 of the first CPU's accesses occur, but see the next point:
495
David Howells6bc39272006-06-25 05:49:22 -0700496 (*) There is no guarantee that a CPU will see the correct order of effects
David Howells108b42b2006-03-31 16:00:29 +0100497 from a second CPU's accesses, even _if_ the second CPU uses a memory
498 barrier, unless the first CPU _also_ uses a matching memory barrier (see
499 the subsection on "SMP Barrier Pairing").
500
501 (*) There is no guarantee that some intervening piece of off-the-CPU
502 hardware[*] will not reorder the memory accesses. CPU cache coherency
503 mechanisms should propagate the indirect effects of a memory barrier
504 between CPUs, but might not do so in order.
505
506 [*] For information on bus mastering DMA and coherency please read:
507
Randy Dunlap4b5ff462008-03-10 17:16:32 -0700508 Documentation/PCI/pci.txt
Paul Bolle395cf962011-08-15 02:02:26 +0200509 Documentation/DMA-API-HOWTO.txt
David Howells108b42b2006-03-31 16:00:29 +0100510 Documentation/DMA-API.txt
511
512
513DATA DEPENDENCY BARRIERS
514------------------------
515
516The usage requirements of data dependency barriers are a little subtle, and
517it's not always obvious that they're needed. To illustrate, consider the
518following sequence of events:
519
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800520 CPU 1 CPU 2
521 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700522 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100523 B = 4;
524 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700525 WRITE_ONCE(P, &B)
526 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800527 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100528
529There's a clear data dependency here, and it would seem that by the end of the
530sequence, Q must be either &A or &B, and that:
531
532 (Q == &A) implies (D == 1)
533 (Q == &B) implies (D == 4)
534
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700535But! CPU 2's perception of P may be updated _before_ its perception of B, thus
David Howells108b42b2006-03-31 16:00:29 +0100536leading to the following situation:
537
538 (Q == &B) and (D == 2) ????
539
540Whilst this may seem like a failure of coherency or causality maintenance, it
541isn't, and this behaviour can be observed on certain real CPUs (such as the DEC
542Alpha).
543
David Howells2b948952006-06-25 05:48:49 -0700544To deal with this, a data dependency barrier or better must be inserted
545between the address load and the data load:
David Howells108b42b2006-03-31 16:00:29 +0100546
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800547 CPU 1 CPU 2
548 =============== ===============
SeongJae Park3dbf0912016-04-12 08:52:52 -0700549 { A == 1, B == 2, C == 3, P == &A, Q == &C }
David Howells108b42b2006-03-31 16:00:29 +0100550 B = 4;
551 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700552 WRITE_ONCE(P, &B);
553 Q = READ_ONCE(P);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800554 <data dependency barrier>
555 D = *Q;
David Howells108b42b2006-03-31 16:00:29 +0100556
557This enforces the occurrence of one of the two implications, and prevents the
558third possibility from arising.
559
Paul E. McKenney92a84dd2016-01-14 14:17:04 -0800560A data-dependency barrier must also order against dependent writes:
561
562 CPU 1 CPU 2
563 =============== ===============
564 { A == 1, B == 2, C = 3, P == &A, Q == &C }
565 B = 4;
566 <write barrier>
567 WRITE_ONCE(P, &B);
568 Q = READ_ONCE(P);
569 <data dependency barrier>
570 *Q = 5;
571
572The data-dependency barrier must order the read into Q with the store
573into *Q. This prohibits this outcome:
574
575 (Q == B) && (B == 4)
576
577Please note that this pattern should be rare. After all, the whole point
578of dependency ordering is to -prevent- writes to the data structure, along
579with the expensive cache misses associated with those writes. This pattern
580can be used to record rare error conditions and the like, and the ordering
581prevents such records from being lost.
582
583
David Howells108b42b2006-03-31 16:00:29 +0100584[!] Note that this extremely counterintuitive situation arises most easily on
585machines with split caches, so that, for example, one cache bank processes
586even-numbered cache lines and the other bank processes odd-numbered cache
587lines. The pointer P might be stored in an odd-numbered cache line, and the
588variable B might be stored in an even-numbered cache line. Then, if the
589even-numbered bank of the reading CPU's cache is extremely busy while the
590odd-numbered bank is idle, one can see the new value of the pointer P (&B),
David Howells6bc39272006-06-25 05:49:22 -0700591but the old value of the variable B (2).
David Howells108b42b2006-03-31 16:00:29 +0100592
593
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800594The data dependency barrier is very important to the RCU system,
595for example. See rcu_assign_pointer() and rcu_dereference() in
596include/linux/rcupdate.h. This permits the current target of an RCU'd
597pointer to be replaced with a new modified target, without the replacement
598target appearing to be incompletely initialised.
David Howells108b42b2006-03-31 16:00:29 +0100599
600See also the subsection on "Cache Coherency" for a more thorough example.
601
602
603CONTROL DEPENDENCIES
604--------------------
605
Paul E. McKenneyff382812015-02-17 10:00:06 -0800606A load-load control dependency requires a full read memory barrier, not
607simply a data dependency barrier to make it work correctly. Consider the
608following bit of code:
David Howells108b42b2006-03-31 16:00:29 +0100609
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700610 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800611 if (q) {
612 <data dependency barrier> /* BUG: No data dependency!!! */
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700613 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700614 }
David Howells108b42b2006-03-31 16:00:29 +0100615
616This will not have the desired effect because there is no actual data
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800617dependency, but rather a control dependency that the CPU may short-circuit
618by attempting to predict the outcome in advance, so that other CPUs see
619the load from b as having happened before the load from a. In such a
620case what's actually required is:
David Howells108b42b2006-03-31 16:00:29 +0100621
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700622 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800623 if (q) {
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700624 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700625 p = READ_ONCE(b);
Paul E. McKenney45c8a362013-07-02 15:24:09 -0700626 }
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800627
628However, stores are not speculated. This means that ordering -is- provided
Paul E. McKenneyff382812015-02-17 10:00:06 -0800629for load-store control dependencies, as in the following example:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800630
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800631 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700632 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700633 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800634 }
635
Paul E. McKenney5af46922015-04-25 12:48:29 -0700636Control dependencies pair normally with other types of barriers. That
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800637said, please note that READ_ONCE() is not optional! Without the
638READ_ONCE(), the compiler might combine the load from 'a' with other
639loads from 'a', and the store to 'b' with other stores to 'b', with
640possible highly counterintuitive effects on ordering.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800641
642Worse yet, if the compiler is able to prove (say) that the value of
643variable 'a' is always non-zero, it would be well within its rights
644to optimize the original example by eliminating the "if" statement
645as follows:
646
647 q = a;
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700648 b = p; /* BUG: Compiler and CPU can both reorder!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800649
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800650So don't leave out the READ_ONCE().
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700651
652It is tempting to try to enforce ordering on identical stores on both
653branches of the "if" statement as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800654
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800655 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800656 if (q) {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800657 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700658 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800659 do_something();
660 } else {
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800661 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700662 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800663 do_something_else();
664 }
665
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700666Unfortunately, current compilers will transform this as follows at high
667optimization levels:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800668
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800669 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700670 barrier();
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700671 WRITE_ONCE(b, p); /* BUG: No ordering vs. load from a!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800672 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700673 /* WRITE_ONCE(b, p); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800674 do_something();
675 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700676 /* WRITE_ONCE(b, p); -- moved up, BUG!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800677 do_something_else();
678 }
679
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700680Now there is no conditional between the load from 'a' and the store to
681'b', which means that the CPU is within its rights to reorder them:
682The conditional is absolutely required, and must be present in the
683assembly code even after all compiler optimizations have been applied.
684Therefore, if you need ordering in this example, you need explicit
685memory barriers, for example, smp_store_release():
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800686
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700687 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700688 if (q) {
689 smp_store_release(&b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800690 do_something();
691 } else {
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700692 smp_store_release(&b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800693 do_something_else();
694 }
695
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700696In contrast, without explicit memory barriers, two-legged-if control
697ordering is guaranteed only when the stores differ, for example:
698
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800699 q = READ_ONCE(a);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700700 if (q) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700701 WRITE_ONCE(b, p);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700702 do_something();
703 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700704 WRITE_ONCE(b, r);
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700705 do_something_else();
706 }
707
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800708The initial READ_ONCE() is still required to prevent the compiler from
709proving the value of 'a'.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800710
711In addition, you need to be careful what you do with the local variable 'q',
712otherwise the compiler might be able to guess the value and again remove
713the needed conditional. For example:
714
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800715 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800716 if (q % MAX) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700717 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800718 do_something();
719 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700720 WRITE_ONCE(b, r);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800721 do_something_else();
722 }
723
724If MAX is defined to be 1, then the compiler knows that (q % MAX) is
725equal to zero, in which case the compiler is within its rights to
726transform the above code into the following:
727
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800728 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700729 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800730 do_something_else();
731
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700732Given this transformation, the CPU is not required to respect the ordering
733between the load from variable 'a' and the store to variable 'b'. It is
734tempting to add a barrier(), but this does not help. The conditional
735is gone, and the barrier won't bring it back. Therefore, if you are
736relying on this ordering, you should make sure that MAX is greater than
737one, perhaps as follows:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800738
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800739 q = READ_ONCE(a);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800740 BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
741 if (q % MAX) {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700742 WRITE_ONCE(b, p);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800743 do_something();
744 } else {
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700745 WRITE_ONCE(b, r);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800746 do_something_else();
747 }
748
Paul E. McKenney2456d2a2014-08-13 15:40:02 -0700749Please note once again that the stores to 'b' differ. If they were
750identical, as noted earlier, the compiler could pull this store outside
751of the 'if' statement.
752
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700753You must also be careful not to rely too much on boolean short-circuit
754evaluation. Consider this example:
755
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800756 q = READ_ONCE(a);
Paul E. McKenney57aecae2015-05-18 18:27:42 -0700757 if (q || 1 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700758 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700759
Paul E. McKenney5af46922015-04-25 12:48:29 -0700760Because the first condition cannot fault and the second condition is
761always true, the compiler can transform this example as following,
762defeating control dependency:
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700763
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800764 q = READ_ONCE(a);
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700765 WRITE_ONCE(b, 1);
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700766
767This example underscores the need to ensure that the compiler cannot
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700768out-guess your code. More generally, although READ_ONCE() does force
Paul E. McKenney8b19d1d2014-10-12 07:55:47 -0700769the compiler to actually emit code for a given load, it does not force
770the compiler to use the results.
771
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800772Finally, control dependencies do -not- provide transitivity. This is
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700773demonstrated by two related examples, with the initial values of
774x and y both being zero:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800775
776 CPU 0 CPU 1
Paul E. McKenney5af46922015-04-25 12:48:29 -0700777 ======================= =======================
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800778 r1 = READ_ONCE(x); r2 = READ_ONCE(y);
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700779 if (r1 > 0) if (r2 > 0)
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700780 WRITE_ONCE(y, 1); WRITE_ONCE(x, 1);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800781
782 assert(!(r1 == 1 && r2 == 1));
783
784The above two-CPU example will never trigger the assert(). However,
785if control dependencies guaranteed transitivity (which they do not),
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700786then adding the following CPU would guarantee a related assertion:
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800787
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700788 CPU 2
789 =====================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700790 WRITE_ONCE(x, 2);
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800791
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700792 assert(!(r1 == 2 && r2 == 1 && x == 2)); /* FAILS!!! */
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800793
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700794But because control dependencies do -not- provide transitivity, the above
795assertion can fail after the combined three-CPU example completes. If you
796need the three-CPU example to provide ordering, you will need smp_mb()
797between the loads and stores in the CPU 0 and CPU 1 code fragments,
Paul E. McKenney5af46922015-04-25 12:48:29 -0700798that is, just before or just after the "if" statements. Furthermore,
799the original two-CPU example is very fragile and should be avoided.
Paul E. McKenney5646f7a2014-07-25 17:05:24 -0700800
801These two examples are the LB and WWC litmus tests from this paper:
802http://www.cl.cam.ac.uk/users/pes20/ppc-supplemental/test6.pdf and this
803site: https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800804
805In summary:
806
807 (*) Control dependencies can order prior loads against later stores.
808 However, they do -not- guarantee any other sort of ordering:
809 Not prior loads against later loads, nor prior stores against
810 later anything. If you need these other forms of ordering,
Davidlohr Buesod87510c2014-12-28 01:11:16 -0800811 use smp_rmb(), smp_wmb(), or, in the case of prior stores and
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800812 later loads, smp_mb().
813
Paul E. McKenney7817b792015-12-29 16:23:18 -0800814 (*) If both legs of the "if" statement begin with identical stores to
815 the same variable, then those stores must be ordered, either by
816 preceding both of them with smp_mb() or by using smp_store_release()
817 to carry out the stores. Please note that it is -not- sufficient
Paul E. McKenneya5052652016-04-12 08:52:49 -0700818 to use barrier() at beginning of each leg of the "if" statement
819 because, as shown by the example above, optimizing compilers can
820 destroy the control dependency while respecting the letter of the
821 barrier() law.
Paul E. McKenney9b2b3bf2014-02-12 20:19:47 -0800822
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800823 (*) Control dependencies require at least one run-time conditional
Paul E. McKenney586dd562014-02-11 12:28:06 -0800824 between the prior load and the subsequent store, and this
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700825 conditional must involve the prior load. If the compiler is able
826 to optimize the conditional away, it will have also optimized
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800827 away the ordering. Careful use of READ_ONCE() and WRITE_ONCE()
828 can help to preserve the needed conditional.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800829
830 (*) Control dependencies require that the compiler avoid reordering the
Linus Torvalds105ff3c2015-11-03 17:22:17 -0800831 dependency into nonexistence. Careful use of READ_ONCE() or
832 atomic{,64}_read() can help to preserve your control dependency.
Paul E. McKenney895f5542016-01-06 14:23:03 -0800833 Please see the COMPILER BARRIER section for more information.
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800834
Paul E. McKenneyff382812015-02-17 10:00:06 -0800835 (*) Control dependencies pair normally with other types of barriers.
836
Peter Zijlstra18c03c62013-12-11 13:59:06 -0800837 (*) Control dependencies do -not- provide transitivity. If you
838 need transitivity, use smp_mb().
David Howells108b42b2006-03-31 16:00:29 +0100839
840
841SMP BARRIER PAIRING
842-------------------
843
844When dealing with CPU-CPU interactions, certain types of memory barrier should
845always be paired. A lack of appropriate pairing is almost certainly an error.
846
Paul E. McKenneyff382812015-02-17 10:00:06 -0800847General barriers pair with each other, though they also pair with most
848other types of barriers, albeit without transitivity. An acquire barrier
849pairs with a release barrier, but both may also pair with other barriers,
850including of course general barriers. A write barrier pairs with a data
851dependency barrier, a control dependency, an acquire barrier, a release
852barrier, a read barrier, or a general barrier. Similarly a read barrier,
853control dependency, or a data dependency barrier pairs with a write
854barrier, an acquire barrier, a release barrier, or a general barrier:
David Howells108b42b2006-03-31 16:00:29 +0100855
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800856 CPU 1 CPU 2
857 =============== ===============
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700858 WRITE_ONCE(a, 1);
David Howells108b42b2006-03-31 16:00:29 +0100859 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700860 WRITE_ONCE(b, 2); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800861 <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700862 y = READ_ONCE(a);
David Howells108b42b2006-03-31 16:00:29 +0100863
864Or:
865
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800866 CPU 1 CPU 2
867 =============== ===============================
David Howells108b42b2006-03-31 16:00:29 +0100868 a = 1;
869 <write barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700870 WRITE_ONCE(b, &a); x = READ_ONCE(b);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800871 <data dependency barrier>
872 y = *x;
David Howells108b42b2006-03-31 16:00:29 +0100873
Paul E. McKenneyff382812015-02-17 10:00:06 -0800874Or even:
875
876 CPU 1 CPU 2
877 =============== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700878 r1 = READ_ONCE(y);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800879 <general barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700880 WRITE_ONCE(y, 1); if (r2 = READ_ONCE(x)) {
Paul E. McKenneyff382812015-02-17 10:00:06 -0800881 <implicit control dependency>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700882 WRITE_ONCE(y, 1);
Paul E. McKenneyff382812015-02-17 10:00:06 -0800883 }
884
885 assert(r1 == 0 || r2 == 0);
886
David Howells108b42b2006-03-31 16:00:29 +0100887Basically, the read barrier always has to be there, even though it can be of
888the "weaker" type.
889
David Howells670bd952006-06-10 09:54:12 -0700890[!] Note that the stores before the write barrier would normally be expected to
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700891match the loads after the read barrier or the data dependency barrier, and vice
David Howells670bd952006-06-10 09:54:12 -0700892versa:
893
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800894 CPU 1 CPU 2
895 =================== ===================
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700896 WRITE_ONCE(a, 1); }---- --->{ v = READ_ONCE(c);
897 WRITE_ONCE(b, 2); } \ / { w = READ_ONCE(d);
Paul E. McKenney2ecf8102013-12-11 13:59:04 -0800898 <write barrier> \ <read barrier>
Paul E. McKenney9af194c2015-06-18 14:33:24 -0700899 WRITE_ONCE(c, 3); } / \ { x = READ_ONCE(a);
900 WRITE_ONCE(d, 4); }---- --->{ y = READ_ONCE(b);
David Howells670bd952006-06-10 09:54:12 -0700901
David Howells108b42b2006-03-31 16:00:29 +0100902
903EXAMPLES OF MEMORY BARRIER SEQUENCES
904------------------------------------
905
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700906Firstly, write barriers act as partial orderings on store operations.
David Howells108b42b2006-03-31 16:00:29 +0100907Consider the following sequence of events:
908
909 CPU 1
910 =======================
911 STORE A = 1
912 STORE B = 2
913 STORE C = 3
914 <write barrier>
915 STORE D = 4
916 STORE E = 5
917
918This sequence of events is committed to the memory coherence system in an order
919that the rest of the system might perceive as the unordered set of { STORE A,
Adrian Bunk80f72282006-06-30 18:27:16 +0200920STORE B, STORE C } all occurring before the unordered set of { STORE D, STORE E
David Howells108b42b2006-03-31 16:00:29 +0100921}:
922
923 +-------+ : :
924 | | +------+
925 | |------>| C=3 | } /\
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700926 | | : +------+ }----- \ -----> Events perceptible to
927 | | : | A=1 | } \/ the rest of the system
David Howells108b42b2006-03-31 16:00:29 +0100928 | | : +------+ }
929 | CPU 1 | : | B=2 | }
930 | | +------+ }
931 | | wwwwwwwwwwwwwwww } <--- At this point the write barrier
932 | | +------+ } requires all stores prior to the
933 | | : | E=5 | } barrier to be committed before
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700934 | | : +------+ } further stores may take place
David Howells108b42b2006-03-31 16:00:29 +0100935 | |------>| D=4 | }
936 | | +------+
937 +-------+ : :
938 |
David Howells670bd952006-06-10 09:54:12 -0700939 | Sequence in which stores are committed to the
940 | memory system by CPU 1
David Howells108b42b2006-03-31 16:00:29 +0100941 V
942
943
Jarek Poplawski81fc6322007-05-23 13:58:20 -0700944Secondly, data dependency barriers act as partial orderings on data-dependent
David Howells108b42b2006-03-31 16:00:29 +0100945loads. Consider the following sequence of events:
946
947 CPU 1 CPU 2
948 ======================= =======================
David Howellsc14038c2006-04-10 22:54:24 -0700949 { B = 7; X = 9; Y = 8; C = &Y }
David Howells108b42b2006-03-31 16:00:29 +0100950 STORE A = 1
951 STORE B = 2
952 <write barrier>
953 STORE C = &B LOAD X
954 STORE D = 4 LOAD C (gets &B)
955 LOAD *C (reads B)
956
957Without intervention, CPU 2 may perceive the events on CPU 1 in some
958effectively random order, despite the write barrier issued by CPU 1:
959
960 +-------+ : : : :
961 | | +------+ +-------+ | Sequence of update
962 | |------>| B=2 |----- --->| Y->8 | | of perception on
963 | | : +------+ \ +-------+ | CPU 2
964 | CPU 1 | : | A=1 | \ --->| C->&Y | V
965 | | +------+ | +-------+
966 | | wwwwwwwwwwwwwwww | : :
967 | | +------+ | : :
968 | | : | C=&B |--- | : : +-------+
969 | | : +------+ \ | +-------+ | |
970 | |------>| D=4 | ----------->| C->&B |------>| |
971 | | +------+ | +-------+ | |
972 +-------+ : : | : : | |
973 | : : | |
974 | : : | CPU 2 |
975 | +-------+ | |
976 Apparently incorrect ---> | | B->7 |------>| |
977 perception of B (!) | +-------+ | |
978 | : : | |
979 | +-------+ | |
980 The load of X holds ---> \ | X->9 |------>| |
981 up the maintenance \ +-------+ | |
982 of coherence of B ----->| B->2 | +-------+
983 +-------+
984 : :
985
986
987In the above example, CPU 2 perceives that B is 7, despite the load of *C
Paolo Ornati670e9f32006-10-03 22:57:56 +0200988(which would be B) coming after the LOAD of C.
David Howells108b42b2006-03-31 16:00:29 +0100989
990If, however, a data dependency barrier were to be placed between the load of C
David Howellsc14038c2006-04-10 22:54:24 -0700991and the load of *C (ie: B) on CPU 2:
992
993 CPU 1 CPU 2
994 ======================= =======================
995 { B = 7; X = 9; Y = 8; C = &Y }
996 STORE A = 1
997 STORE B = 2
998 <write barrier>
999 STORE C = &B LOAD X
1000 STORE D = 4 LOAD C (gets &B)
1001 <data dependency barrier>
1002 LOAD *C (reads B)
1003
1004then the following will occur:
David Howells108b42b2006-03-31 16:00:29 +01001005
1006 +-------+ : : : :
1007 | | +------+ +-------+
1008 | |------>| B=2 |----- --->| Y->8 |
1009 | | : +------+ \ +-------+
1010 | CPU 1 | : | A=1 | \ --->| C->&Y |
1011 | | +------+ | +-------+
1012 | | wwwwwwwwwwwwwwww | : :
1013 | | +------+ | : :
1014 | | : | C=&B |--- | : : +-------+
1015 | | : +------+ \ | +-------+ | |
1016 | |------>| D=4 | ----------->| C->&B |------>| |
1017 | | +------+ | +-------+ | |
1018 +-------+ : : | : : | |
1019 | : : | |
1020 | : : | CPU 2 |
1021 | +-------+ | |
David Howells670bd952006-06-10 09:54:12 -07001022 | | X->9 |------>| |
1023 | +-------+ | |
1024 Makes sure all effects ---> \ ddddddddddddddddd | |
1025 prior to the store of C \ +-------+ | |
1026 are perceptible to ----->| B->2 |------>| |
1027 subsequent loads +-------+ | |
David Howells108b42b2006-03-31 16:00:29 +01001028 : : +-------+
1029
1030
1031And thirdly, a read barrier acts as a partial order on loads. Consider the
1032following sequence of events:
1033
1034 CPU 1 CPU 2
1035 ======================= =======================
David Howells670bd952006-06-10 09:54:12 -07001036 { A = 0, B = 9 }
David Howells108b42b2006-03-31 16:00:29 +01001037 STORE A=1
David Howells108b42b2006-03-31 16:00:29 +01001038 <write barrier>
David Howells670bd952006-06-10 09:54:12 -07001039 STORE B=2
David Howells108b42b2006-03-31 16:00:29 +01001040 LOAD B
David Howells670bd952006-06-10 09:54:12 -07001041 LOAD A
David Howells108b42b2006-03-31 16:00:29 +01001042
1043Without intervention, CPU 2 may then choose to perceive the events on CPU 1 in
1044some effectively random order, despite the write barrier issued by CPU 1:
1045
David Howells670bd952006-06-10 09:54:12 -07001046 +-------+ : : : :
1047 | | +------+ +-------+
1048 | |------>| A=1 |------ --->| A->0 |
1049 | | +------+ \ +-------+
1050 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1051 | | +------+ | +-------+
1052 | |------>| B=2 |--- | : :
1053 | | +------+ \ | : : +-------+
1054 +-------+ : : \ | +-------+ | |
1055 ---------->| B->2 |------>| |
1056 | +-------+ | CPU 2 |
1057 | | A->0 |------>| |
1058 | +-------+ | |
1059 | : : +-------+
1060 \ : :
1061 \ +-------+
1062 ---->| A->1 |
1063 +-------+
1064 : :
David Howells108b42b2006-03-31 16:00:29 +01001065
1066
David Howells6bc39272006-06-25 05:49:22 -07001067If, however, a read barrier were to be placed between the load of B and the
David Howells670bd952006-06-10 09:54:12 -07001068load of A on CPU 2:
David Howells108b42b2006-03-31 16:00:29 +01001069
David Howells670bd952006-06-10 09:54:12 -07001070 CPU 1 CPU 2
1071 ======================= =======================
1072 { A = 0, B = 9 }
1073 STORE A=1
1074 <write barrier>
1075 STORE B=2
1076 LOAD B
1077 <read barrier>
1078 LOAD A
1079
1080then the partial ordering imposed by CPU 1 will be perceived correctly by CPU
10812:
1082
1083 +-------+ : : : :
1084 | | +------+ +-------+
1085 | |------>| A=1 |------ --->| A->0 |
1086 | | +------+ \ +-------+
1087 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1088 | | +------+ | +-------+
1089 | |------>| B=2 |--- | : :
1090 | | +------+ \ | : : +-------+
1091 +-------+ : : \ | +-------+ | |
1092 ---------->| B->2 |------>| |
1093 | +-------+ | CPU 2 |
1094 | : : | |
1095 | : : | |
1096 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1097 barrier causes all effects \ +-------+ | |
1098 prior to the storage of B ---->| A->1 |------>| |
1099 to be perceptible to CPU 2 +-------+ | |
1100 : : +-------+
1101
1102
1103To illustrate this more completely, consider what could happen if the code
1104contained a load of A either side of the read barrier:
1105
1106 CPU 1 CPU 2
1107 ======================= =======================
1108 { A = 0, B = 9 }
1109 STORE A=1
1110 <write barrier>
1111 STORE B=2
1112 LOAD B
1113 LOAD A [first load of A]
1114 <read barrier>
1115 LOAD A [second load of A]
1116
1117Even though the two loads of A both occur after the load of B, they may both
1118come up with different values:
1119
1120 +-------+ : : : :
1121 | | +------+ +-------+
1122 | |------>| A=1 |------ --->| A->0 |
1123 | | +------+ \ +-------+
1124 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1125 | | +------+ | +-------+
1126 | |------>| B=2 |--- | : :
1127 | | +------+ \ | : : +-------+
1128 +-------+ : : \ | +-------+ | |
1129 ---------->| B->2 |------>| |
1130 | +-------+ | CPU 2 |
1131 | : : | |
1132 | : : | |
1133 | +-------+ | |
1134 | | A->0 |------>| 1st |
1135 | +-------+ | |
1136 At this point the read ----> \ rrrrrrrrrrrrrrrrr | |
1137 barrier causes all effects \ +-------+ | |
1138 prior to the storage of B ---->| A->1 |------>| 2nd |
1139 to be perceptible to CPU 2 +-------+ | |
1140 : : +-------+
1141
1142
1143But it may be that the update to A from CPU 1 becomes perceptible to CPU 2
1144before the read barrier completes anyway:
1145
1146 +-------+ : : : :
1147 | | +------+ +-------+
1148 | |------>| A=1 |------ --->| A->0 |
1149 | | +------+ \ +-------+
1150 | CPU 1 | wwwwwwwwwwwwwwww \ --->| B->9 |
1151 | | +------+ | +-------+
1152 | |------>| B=2 |--- | : :
1153 | | +------+ \ | : : +-------+
1154 +-------+ : : \ | +-------+ | |
1155 ---------->| B->2 |------>| |
1156 | +-------+ | CPU 2 |
1157 | : : | |
1158 \ : : | |
1159 \ +-------+ | |
1160 ---->| A->1 |------>| 1st |
1161 +-------+ | |
1162 rrrrrrrrrrrrrrrrr | |
1163 +-------+ | |
1164 | A->1 |------>| 2nd |
1165 +-------+ | |
1166 : : +-------+
1167
1168
1169The guarantee is that the second load will always come up with A == 1 if the
1170load of B came up with B == 2. No such guarantee exists for the first load of
1171A; that may come up with either A == 0 or A == 1.
1172
1173
1174READ MEMORY BARRIERS VS LOAD SPECULATION
1175----------------------------------------
1176
1177Many CPUs speculate with loads: that is they see that they will need to load an
1178item from memory, and they find a time where they're not using the bus for any
1179other loads, and so do the load in advance - even though they haven't actually
1180got to that point in the instruction execution flow yet. This permits the
1181actual load instruction to potentially complete immediately because the CPU
1182already has the value to hand.
1183
1184It may turn out that the CPU didn't actually need the value - perhaps because a
1185branch circumvented the load - in which case it can discard the value or just
1186cache it for later use.
1187
1188Consider:
1189
Ingo Molnare0edc782013-11-22 11:24:53 +01001190 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001191 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001192 LOAD B
1193 DIVIDE } Divide instructions generally
1194 DIVIDE } take a long time to perform
1195 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001196
1197Which might appear as this:
1198
1199 : : +-------+
1200 +-------+ | |
1201 --->| B->2 |------>| |
1202 +-------+ | CPU 2 |
1203 : :DIVIDE | |
1204 +-------+ | |
1205 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1206 division speculates on the +-------+ ~ | |
1207 LOAD of A : : ~ | |
1208 : :DIVIDE | |
1209 : : ~ | |
1210 Once the divisions are complete --> : : ~-->| |
1211 the CPU can then perform the : : | |
1212 LOAD with immediate effect : : +-------+
1213
1214
1215Placing a read barrier or a data dependency barrier just before the second
1216load:
1217
Ingo Molnare0edc782013-11-22 11:24:53 +01001218 CPU 1 CPU 2
David Howells670bd952006-06-10 09:54:12 -07001219 ======================= =======================
Ingo Molnare0edc782013-11-22 11:24:53 +01001220 LOAD B
1221 DIVIDE
1222 DIVIDE
David Howells670bd952006-06-10 09:54:12 -07001223 <read barrier>
Ingo Molnare0edc782013-11-22 11:24:53 +01001224 LOAD A
David Howells670bd952006-06-10 09:54:12 -07001225
1226will force any value speculatively obtained to be reconsidered to an extent
1227dependent on the type of barrier used. If there was no change made to the
1228speculated memory location, then the speculated value will just be used:
1229
1230 : : +-------+
1231 +-------+ | |
1232 --->| B->2 |------>| |
1233 +-------+ | CPU 2 |
1234 : :DIVIDE | |
1235 +-------+ | |
1236 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1237 division speculates on the +-------+ ~ | |
1238 LOAD of A : : ~ | |
1239 : :DIVIDE | |
1240 : : ~ | |
1241 : : ~ | |
1242 rrrrrrrrrrrrrrrr~ | |
1243 : : ~ | |
1244 : : ~-->| |
1245 : : | |
1246 : : +-------+
1247
1248
1249but if there was an update or an invalidation from another CPU pending, then
1250the speculation will be cancelled and the value reloaded:
1251
1252 : : +-------+
1253 +-------+ | |
1254 --->| B->2 |------>| |
1255 +-------+ | CPU 2 |
1256 : :DIVIDE | |
1257 +-------+ | |
1258 The CPU being busy doing a ---> --->| A->0 |~~~~ | |
1259 division speculates on the +-------+ ~ | |
1260 LOAD of A : : ~ | |
1261 : :DIVIDE | |
1262 : : ~ | |
1263 : : ~ | |
1264 rrrrrrrrrrrrrrrrr | |
1265 +-------+ | |
1266 The speculation is discarded ---> --->| A->1 |------>| |
1267 and an updated value is +-------+ | |
1268 retrieved : : +-------+
David Howells108b42b2006-03-31 16:00:29 +01001269
1270
Paul E. McKenney241e6662011-02-10 16:54:50 -08001271TRANSITIVITY
1272------------
1273
1274Transitivity is a deeply intuitive notion about ordering that is not
1275always provided by real computer systems. The following example
Paul E. McKenneyf36fe1e2016-02-15 14:50:36 -08001276demonstrates transitivity:
Paul E. McKenney241e6662011-02-10 16:54:50 -08001277
1278 CPU 1 CPU 2 CPU 3
1279 ======================= ======================= =======================
1280 { X = 0, Y = 0 }
1281 STORE X=1 LOAD X STORE Y=1
1282 <general barrier> <general barrier>
1283 LOAD Y LOAD X
1284
1285Suppose that CPU 2's load from X returns 1 and its load from Y returns 0.
1286This indicates that CPU 2's load from X in some sense follows CPU 1's
1287store to X and that CPU 2's load from Y in some sense preceded CPU 3's
1288store to Y. The question is then "Can CPU 3's load from X return 0?"
1289
1290Because CPU 2's load from X in some sense came after CPU 1's store, it
1291is natural to expect that CPU 3's load from X must therefore return 1.
1292This expectation is an example of transitivity: if a load executing on
1293CPU A follows a load from the same variable executing on CPU B, then
1294CPU A's load must either return the same value that CPU B's load did,
1295or must return some later value.
1296
1297In the Linux kernel, use of general memory barriers guarantees
1298transitivity. Therefore, in the above example, if CPU 2's load from X
1299returns 1 and its load from Y returns 0, then CPU 3's load from X must
1300also return 1.
1301
1302However, transitivity is -not- guaranteed for read or write barriers.
1303For example, suppose that CPU 2's general barrier in the above example
1304is changed to a read barrier as shown below:
1305
1306 CPU 1 CPU 2 CPU 3
1307 ======================= ======================= =======================
1308 { X = 0, Y = 0 }
1309 STORE X=1 LOAD X STORE Y=1
1310 <read barrier> <general barrier>
1311 LOAD Y LOAD X
1312
1313This substitution destroys transitivity: in this example, it is perfectly
1314legal for CPU 2's load from X to return 1, its load from Y to return 0,
1315and CPU 3's load from X to return 0.
1316
1317The key point is that although CPU 2's read barrier orders its pair
1318of loads, it does not guarantee to order CPU 1's store. Therefore, if
1319this example runs on a system where CPUs 1 and 2 share a store buffer
1320or a level of cache, CPU 2 might have early access to CPU 1's writes.
1321General barriers are therefore required to ensure that all CPUs agree
1322on the combined order of CPU 1's and CPU 2's accesses.
1323
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001324General barriers provide "global transitivity", so that all CPUs will
1325agree on the order of operations. In contrast, a chain of release-acquire
1326pairs provides only "local transitivity", so that only those CPUs on
1327the chain are guaranteed to agree on the combined order of the accesses.
1328For example, switching to C code in deference to Herman Hollerith:
1329
1330 int u, v, x, y, z;
1331
1332 void cpu0(void)
1333 {
1334 r0 = smp_load_acquire(&x);
1335 WRITE_ONCE(u, 1);
1336 smp_store_release(&y, 1);
1337 }
1338
1339 void cpu1(void)
1340 {
1341 r1 = smp_load_acquire(&y);
1342 r4 = READ_ONCE(v);
1343 r5 = READ_ONCE(u);
1344 smp_store_release(&z, 1);
1345 }
1346
1347 void cpu2(void)
1348 {
1349 r2 = smp_load_acquire(&z);
1350 smp_store_release(&x, 1);
1351 }
1352
1353 void cpu3(void)
1354 {
1355 WRITE_ONCE(v, 1);
1356 smp_mb();
1357 r3 = READ_ONCE(u);
1358 }
1359
1360Because cpu0(), cpu1(), and cpu2() participate in a local transitive
1361chain of smp_store_release()/smp_load_acquire() pairs, the following
1362outcome is prohibited:
1363
1364 r0 == 1 && r1 == 1 && r2 == 1
1365
1366Furthermore, because of the release-acquire relationship between cpu0()
1367and cpu1(), cpu1() must see cpu0()'s writes, so that the following
1368outcome is prohibited:
1369
1370 r1 == 1 && r5 == 0
1371
1372However, the transitivity of release-acquire is local to the participating
1373CPUs and does not apply to cpu3(). Therefore, the following outcome
1374is possible:
1375
1376 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0
1377
Paul E. McKenney37ef0342016-01-25 22:12:34 -08001378As an aside, the following outcome is also possible:
1379
1380 r0 == 0 && r1 == 1 && r2 == 1 && r3 == 0 && r4 == 0 && r5 == 1
1381
Paul E. McKenneyc535cc92016-01-15 09:30:42 -08001382Although cpu0(), cpu1(), and cpu2() will see their respective reads and
1383writes in order, CPUs not involved in the release-acquire chain might
1384well disagree on the order. This disagreement stems from the fact that
1385the weak memory-barrier instructions used to implement smp_load_acquire()
1386and smp_store_release() are not required to order prior stores against
1387subsequent loads in all cases. This means that cpu3() can see cpu0()'s
1388store to u as happening -after- cpu1()'s load from v, even though
1389both cpu0() and cpu1() agree that these two operations occurred in the
1390intended order.
1391
1392However, please keep in mind that smp_load_acquire() is not magic.
1393In particular, it simply reads from its argument with ordering. It does
1394-not- ensure that any particular value will be read. Therefore, the
1395following outcome is possible:
1396
1397 r0 == 0 && r1 == 0 && r2 == 0 && r5 == 0
1398
1399Note that this outcome can happen even on a mythical sequentially
1400consistent system where nothing is ever reordered.
1401
1402To reiterate, if your code requires global transitivity, use general
1403barriers throughout.
Paul E. McKenney241e6662011-02-10 16:54:50 -08001404
1405
David Howells108b42b2006-03-31 16:00:29 +01001406========================
1407EXPLICIT KERNEL BARRIERS
1408========================
1409
1410The Linux kernel has a variety of different barriers that act at different
1411levels:
1412
1413 (*) Compiler barrier.
1414
1415 (*) CPU memory barriers.
1416
1417 (*) MMIO write barrier.
1418
1419
1420COMPILER BARRIER
1421----------------
1422
1423The Linux kernel has an explicit compiler barrier function that prevents the
1424compiler from moving the memory accesses either side of it to the other side:
1425
1426 barrier();
1427
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001428This is a general barrier -- there are no read-read or write-write
1429variants of barrier(). However, READ_ONCE() and WRITE_ONCE() can be
1430thought of as weak forms of barrier() that affect only the specific
1431accesses flagged by the READ_ONCE() or WRITE_ONCE().
David Howells108b42b2006-03-31 16:00:29 +01001432
Paul E. McKenney692118d2013-12-11 13:59:07 -08001433The barrier() function has the following effects:
1434
1435 (*) Prevents the compiler from reordering accesses following the
1436 barrier() to precede any accesses preceding the barrier().
1437 One example use for this property is to ease communication between
1438 interrupt-handler code and the code that was interrupted.
1439
1440 (*) Within a loop, forces the compiler to load the variables used
1441 in that loop's conditional on each pass through that loop.
1442
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001443The READ_ONCE() and WRITE_ONCE() functions can prevent any number of
1444optimizations that, while perfectly safe in single-threaded code, can
1445be fatal in concurrent code. Here are some examples of these sorts
1446of optimizations:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001447
Paul E. McKenney449f7412014-01-02 15:03:50 -08001448 (*) The compiler is within its rights to reorder loads and stores
1449 to the same variable, and in some cases, the CPU is within its
1450 rights to reorder loads to the same variable. This means that
1451 the following code:
1452
1453 a[0] = x;
1454 a[1] = x;
1455
1456 Might result in an older value of x stored in a[1] than in a[0].
1457 Prevent both the compiler and the CPU from doing this as follows:
1458
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001459 a[0] = READ_ONCE(x);
1460 a[1] = READ_ONCE(x);
Paul E. McKenney449f7412014-01-02 15:03:50 -08001461
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001462 In short, READ_ONCE() and WRITE_ONCE() provide cache coherence for
1463 accesses from multiple CPUs to a single variable.
Paul E. McKenney449f7412014-01-02 15:03:50 -08001464
Paul E. McKenney692118d2013-12-11 13:59:07 -08001465 (*) The compiler is within its rights to merge successive loads from
1466 the same variable. Such merging can cause the compiler to "optimize"
1467 the following code:
1468
1469 while (tmp = a)
1470 do_something_with(tmp);
1471
1472 into the following code, which, although in some sense legitimate
1473 for single-threaded code, is almost certainly not what the developer
1474 intended:
1475
1476 if (tmp = a)
1477 for (;;)
1478 do_something_with(tmp);
1479
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001480 Use READ_ONCE() to prevent the compiler from doing this to you:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001481
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001482 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001483 do_something_with(tmp);
1484
1485 (*) The compiler is within its rights to reload a variable, for example,
1486 in cases where high register pressure prevents the compiler from
1487 keeping all data of interest in registers. The compiler might
1488 therefore optimize the variable 'tmp' out of our previous example:
1489
1490 while (tmp = a)
1491 do_something_with(tmp);
1492
1493 This could result in the following code, which is perfectly safe in
1494 single-threaded code, but can be fatal in concurrent code:
1495
1496 while (a)
1497 do_something_with(a);
1498
1499 For example, the optimized version of this code could result in
1500 passing a zero to do_something_with() in the case where the variable
1501 a was modified by some other CPU between the "while" statement and
1502 the call to do_something_with().
1503
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001504 Again, use READ_ONCE() to prevent the compiler from doing this:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001505
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001506 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001507 do_something_with(tmp);
1508
1509 Note that if the compiler runs short of registers, it might save
1510 tmp onto the stack. The overhead of this saving and later restoring
1511 is why compilers reload variables. Doing so is perfectly safe for
1512 single-threaded code, so you need to tell the compiler about cases
1513 where it is not safe.
1514
1515 (*) The compiler is within its rights to omit a load entirely if it knows
1516 what the value will be. For example, if the compiler can prove that
1517 the value of variable 'a' is always zero, it can optimize this code:
1518
1519 while (tmp = a)
1520 do_something_with(tmp);
1521
1522 Into this:
1523
1524 do { } while (0);
1525
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001526 This transformation is a win for single-threaded code because it
1527 gets rid of a load and a branch. The problem is that the compiler
1528 will carry out its proof assuming that the current CPU is the only
1529 one updating variable 'a'. If variable 'a' is shared, then the
1530 compiler's proof will be erroneous. Use READ_ONCE() to tell the
1531 compiler that it doesn't know as much as it thinks it does:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001532
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001533 while (tmp = READ_ONCE(a))
Paul E. McKenney692118d2013-12-11 13:59:07 -08001534 do_something_with(tmp);
1535
1536 But please note that the compiler is also closely watching what you
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001537 do with the value after the READ_ONCE(). For example, suppose you
Paul E. McKenney692118d2013-12-11 13:59:07 -08001538 do the following and MAX is a preprocessor macro with the value 1:
1539
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001540 while ((tmp = READ_ONCE(a)) % MAX)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001541 do_something_with(tmp);
1542
1543 Then the compiler knows that the result of the "%" operator applied
1544 to MAX will always be zero, again allowing the compiler to optimize
1545 the code into near-nonexistence. (It will still load from the
1546 variable 'a'.)
1547
1548 (*) Similarly, the compiler is within its rights to omit a store entirely
1549 if it knows that the variable already has the value being stored.
1550 Again, the compiler assumes that the current CPU is the only one
1551 storing into the variable, which can cause the compiler to do the
1552 wrong thing for shared variables. For example, suppose you have
1553 the following:
1554
1555 a = 0;
SeongJae Park65f95ff2016-02-22 08:28:29 -08001556 ... Code that does not store to variable a ...
Paul E. McKenney692118d2013-12-11 13:59:07 -08001557 a = 0;
1558
1559 The compiler sees that the value of variable 'a' is already zero, so
1560 it might well omit the second store. This would come as a fatal
1561 surprise if some other CPU might have stored to variable 'a' in the
1562 meantime.
1563
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001564 Use WRITE_ONCE() to prevent the compiler from making this sort of
Paul E. McKenney692118d2013-12-11 13:59:07 -08001565 wrong guess:
1566
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001567 WRITE_ONCE(a, 0);
SeongJae Park65f95ff2016-02-22 08:28:29 -08001568 ... Code that does not store to variable a ...
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001569 WRITE_ONCE(a, 0);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001570
1571 (*) The compiler is within its rights to reorder memory accesses unless
1572 you tell it not to. For example, consider the following interaction
1573 between process-level code and an interrupt handler:
1574
1575 void process_level(void)
1576 {
1577 msg = get_message();
1578 flag = true;
1579 }
1580
1581 void interrupt_handler(void)
1582 {
1583 if (flag)
1584 process_message(msg);
1585 }
1586
Masanari Iidadf5cbb22014-03-21 10:04:30 +09001587 There is nothing to prevent the compiler from transforming
Paul E. McKenney692118d2013-12-11 13:59:07 -08001588 process_level() to the following, in fact, this might well be a
1589 win for single-threaded code:
1590
1591 void process_level(void)
1592 {
1593 flag = true;
1594 msg = get_message();
1595 }
1596
1597 If the interrupt occurs between these two statement, then
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001598 interrupt_handler() might be passed a garbled msg. Use WRITE_ONCE()
Paul E. McKenney692118d2013-12-11 13:59:07 -08001599 to prevent this as follows:
1600
1601 void process_level(void)
1602 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001603 WRITE_ONCE(msg, get_message());
1604 WRITE_ONCE(flag, true);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001605 }
1606
1607 void interrupt_handler(void)
1608 {
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001609 if (READ_ONCE(flag))
1610 process_message(READ_ONCE(msg));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001611 }
1612
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001613 Note that the READ_ONCE() and WRITE_ONCE() wrappers in
1614 interrupt_handler() are needed if this interrupt handler can itself
1615 be interrupted by something that also accesses 'flag' and 'msg',
1616 for example, a nested interrupt or an NMI. Otherwise, READ_ONCE()
1617 and WRITE_ONCE() are not needed in interrupt_handler() other than
1618 for documentation purposes. (Note also that nested interrupts
1619 do not typically occur in modern Linux kernels, in fact, if an
1620 interrupt handler returns with interrupts enabled, you will get a
1621 WARN_ONCE() splat.)
Paul E. McKenney692118d2013-12-11 13:59:07 -08001622
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001623 You should assume that the compiler can move READ_ONCE() and
1624 WRITE_ONCE() past code not containing READ_ONCE(), WRITE_ONCE(),
1625 barrier(), or similar primitives.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001626
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001627 This effect could also be achieved using barrier(), but READ_ONCE()
1628 and WRITE_ONCE() are more selective: With READ_ONCE() and
1629 WRITE_ONCE(), the compiler need only forget the contents of the
1630 indicated memory locations, while with barrier() the compiler must
1631 discard the value of all memory locations that it has currented
1632 cached in any machine registers. Of course, the compiler must also
1633 respect the order in which the READ_ONCE()s and WRITE_ONCE()s occur,
1634 though the CPU of course need not do so.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001635
1636 (*) The compiler is within its rights to invent stores to a variable,
1637 as in the following example:
1638
1639 if (a)
1640 b = a;
1641 else
1642 b = 42;
1643
1644 The compiler might save a branch by optimizing this as follows:
1645
1646 b = 42;
1647 if (a)
1648 b = a;
1649
1650 In single-threaded code, this is not only safe, but also saves
1651 a branch. Unfortunately, in concurrent code, this optimization
1652 could cause some other CPU to see a spurious value of 42 -- even
1653 if variable 'a' was never zero -- when loading variable 'b'.
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001654 Use WRITE_ONCE() to prevent this as follows:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001655
1656 if (a)
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001657 WRITE_ONCE(b, a);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001658 else
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001659 WRITE_ONCE(b, 42);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001660
1661 The compiler can also invent loads. These are usually less
1662 damaging, but they can result in cache-line bouncing and thus in
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001663 poor performance and scalability. Use READ_ONCE() to prevent
Paul E. McKenney692118d2013-12-11 13:59:07 -08001664 invented loads.
1665
1666 (*) For aligned memory locations whose size allows them to be accessed
1667 with a single memory-reference instruction, prevents "load tearing"
1668 and "store tearing," in which a single large access is replaced by
1669 multiple smaller accesses. For example, given an architecture having
1670 16-bit store instructions with 7-bit immediate fields, the compiler
1671 might be tempted to use two 16-bit store-immediate instructions to
1672 implement the following 32-bit store:
1673
1674 p = 0x00010002;
1675
1676 Please note that GCC really does use this sort of optimization,
1677 which is not surprising given that it would likely take more
1678 than two instructions to build the constant and then store it.
1679 This optimization can therefore be a win in single-threaded code.
1680 In fact, a recent bug (since fixed) caused GCC to incorrectly use
1681 this optimization in a volatile store. In the absence of such bugs,
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001682 use of WRITE_ONCE() prevents store tearing in the following example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001683
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001684 WRITE_ONCE(p, 0x00010002);
Paul E. McKenney692118d2013-12-11 13:59:07 -08001685
1686 Use of packed structures can also result in load and store tearing,
1687 as in this example:
1688
1689 struct __attribute__((__packed__)) foo {
1690 short a;
1691 int b;
1692 short c;
1693 };
1694 struct foo foo1, foo2;
1695 ...
1696
1697 foo2.a = foo1.a;
1698 foo2.b = foo1.b;
1699 foo2.c = foo1.c;
1700
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001701 Because there are no READ_ONCE() or WRITE_ONCE() wrappers and no
1702 volatile markings, the compiler would be well within its rights to
1703 implement these three assignment statements as a pair of 32-bit
1704 loads followed by a pair of 32-bit stores. This would result in
1705 load tearing on 'foo1.b' and store tearing on 'foo2.b'. READ_ONCE()
1706 and WRITE_ONCE() again prevent tearing in this example:
Paul E. McKenney692118d2013-12-11 13:59:07 -08001707
1708 foo2.a = foo1.a;
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001709 WRITE_ONCE(foo2.b, READ_ONCE(foo1.b));
Paul E. McKenney692118d2013-12-11 13:59:07 -08001710 foo2.c = foo1.c;
1711
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001712All that aside, it is never necessary to use READ_ONCE() and
1713WRITE_ONCE() on a variable that has been marked volatile. For example,
1714because 'jiffies' is marked volatile, it is never necessary to
1715say READ_ONCE(jiffies). The reason for this is that READ_ONCE() and
1716WRITE_ONCE() are implemented as volatile casts, which has no effect when
1717its argument is already marked volatile.
Paul E. McKenney692118d2013-12-11 13:59:07 -08001718
1719Please note that these compiler barriers have no direct effect on the CPU,
1720which may then reorder things however it wishes.
David Howells108b42b2006-03-31 16:00:29 +01001721
1722
1723CPU MEMORY BARRIERS
1724-------------------
1725
1726The Linux kernel has eight basic CPU memory barriers:
1727
1728 TYPE MANDATORY SMP CONDITIONAL
1729 =============== ======================= ===========================
1730 GENERAL mb() smp_mb()
1731 WRITE wmb() smp_wmb()
1732 READ rmb() smp_rmb()
1733 DATA DEPENDENCY read_barrier_depends() smp_read_barrier_depends()
1734
1735
Nick Piggin73f10282008-05-14 06:35:11 +02001736All memory barriers except the data dependency barriers imply a compiler
SeongJae Park0b6fa342016-04-12 08:52:53 -07001737barrier. Data dependencies do not impose any additional compiler ordering.
Nick Piggin73f10282008-05-14 06:35:11 +02001738
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001739Aside: In the case of data dependencies, the compiler would be expected
1740to issue the loads in the correct order (eg. `a[b]` would have to load
1741the value of b before loading a[b]), however there is no guarantee in
1742the C specification that the compiler may not speculate the value of b
1743(eg. is equal to 1) and load a before b (eg. tmp = a[1]; if (b != 1)
SeongJae Park0b6fa342016-04-12 08:52:53 -07001744tmp = a[b]; ). There is also the problem of a compiler reloading b after
1745having loaded a[b], thus having a newer copy of b than a[b]. A consensus
Paul E. McKenney9af194c2015-06-18 14:33:24 -07001746has not yet been reached about these problems, however the READ_ONCE()
1747macro is a good place to start looking.
David Howells108b42b2006-03-31 16:00:29 +01001748
1749SMP memory barriers are reduced to compiler barriers on uniprocessor compiled
Jarek Poplawski81fc6322007-05-23 13:58:20 -07001750systems because it is assumed that a CPU will appear to be self-consistent,
David Howells108b42b2006-03-31 16:00:29 +01001751and will order overlapping accesses correctly with respect to itself.
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001752However, see the subsection on "Virtual Machine Guests" below.
David Howells108b42b2006-03-31 16:00:29 +01001753
1754[!] Note that SMP memory barriers _must_ be used to control the ordering of
1755references to shared memory on SMP systems, though the use of locking instead
1756is sufficient.
1757
1758Mandatory barriers should not be used to control SMP effects, since mandatory
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02001759barriers impose unnecessary overhead on both SMP and UP systems. They may,
1760however, be used to control MMIO effects on accesses through relaxed memory I/O
1761windows. These barriers are required even on non-SMP systems as they affect
1762the order in which memory operations appear to a device by prohibiting both the
1763compiler and the CPU from reordering them.
David Howells108b42b2006-03-31 16:00:29 +01001764
1765
1766There are some more advanced barrier functions:
1767
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02001768 (*) smp_store_mb(var, value)
David Howells108b42b2006-03-31 16:00:29 +01001769
Oleg Nesterov75b2bd52006-11-08 17:44:38 -08001770 This assigns the value to the variable and then inserts a full memory
Davidlohr Bueso2d142e52015-10-27 12:53:51 -07001771 barrier after it. It isn't guaranteed to insert anything more than a
1772 compiler barrier in a UP compilation.
David Howells108b42b2006-03-31 16:00:29 +01001773
1774
Peter Zijlstra1b156112014-03-13 19:00:35 +01001775 (*) smp_mb__before_atomic();
1776 (*) smp_mb__after_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001777
Peter Zijlstra1b156112014-03-13 19:00:35 +01001778 These are for use with atomic (such as add, subtract, increment and
1779 decrement) functions that don't return a value, especially when used for
1780 reference counting. These functions do not imply memory barriers.
1781
1782 These are also used for atomic bitop functions that do not return a
1783 value (such as set_bit and clear_bit).
David Howells108b42b2006-03-31 16:00:29 +01001784
1785 As an example, consider a piece of code that marks an object as being dead
1786 and then decrements the object's reference count:
1787
1788 obj->dead = 1;
Peter Zijlstra1b156112014-03-13 19:00:35 +01001789 smp_mb__before_atomic();
David Howells108b42b2006-03-31 16:00:29 +01001790 atomic_dec(&obj->ref_count);
1791
1792 This makes sure that the death mark on the object is perceived to be set
1793 *before* the reference counter is decremented.
1794
1795 See Documentation/atomic_ops.txt for more information. See the "Atomic
1796 operations" subsection for information on where to use these.
1797
1798
Paul E. McKenneyad2ad5d2015-09-17 08:18:32 -07001799 (*) lockless_dereference();
SeongJae Park0b6fa342016-04-12 08:52:53 -07001800
Paul E. McKenneyad2ad5d2015-09-17 08:18:32 -07001801 This can be thought of as a pointer-fetch wrapper around the
1802 smp_read_barrier_depends() data-dependency barrier.
1803
1804 This is also similar to rcu_dereference(), but in cases where
1805 object lifetime is handled by some mechanism other than RCU, for
1806 example, when the objects removed only when the system goes down.
1807 In addition, lockless_dereference() is used in some data structures
1808 that can be used both with and without RCU.
1809
1810
Alexander Duyck1077fa32014-12-11 15:02:06 -08001811 (*) dma_wmb();
1812 (*) dma_rmb();
1813
1814 These are for use with consistent memory to guarantee the ordering
1815 of writes or reads of shared memory accessible to both the CPU and a
1816 DMA capable device.
1817
1818 For example, consider a device driver that shares memory with a device
1819 and uses a descriptor status value to indicate if the descriptor belongs
1820 to the device or the CPU, and a doorbell to notify it when new
1821 descriptors are available:
1822
1823 if (desc->status != DEVICE_OWN) {
1824 /* do not read data until we own descriptor */
1825 dma_rmb();
1826
1827 /* read/modify data */
1828 read_data = desc->data;
1829 desc->data = write_data;
1830
1831 /* flush modifications before status update */
1832 dma_wmb();
1833
1834 /* assign ownership */
1835 desc->status = DEVICE_OWN;
1836
1837 /* force memory to sync before notifying device via MMIO */
1838 wmb();
1839
1840 /* notify device of new descriptors */
1841 writel(DESC_NOTIFY, doorbell);
1842 }
1843
1844 The dma_rmb() allows us guarantee the device has released ownership
Sylvain Trias7a458002015-04-08 10:27:57 +02001845 before we read the data from the descriptor, and the dma_wmb() allows
Alexander Duyck1077fa32014-12-11 15:02:06 -08001846 us to guarantee the data is written to the descriptor before the device
1847 can see it now has ownership. The wmb() is needed to guarantee that the
1848 cache coherent memory writes have completed before attempting a write to
1849 the cache incoherent MMIO region.
1850
1851 See Documentation/DMA-API.txt for more information on consistent memory.
1852
David Howells108b42b2006-03-31 16:00:29 +01001853MMIO WRITE BARRIER
1854------------------
1855
1856The Linux kernel also has a special barrier for use with memory-mapped I/O
1857writes:
1858
1859 mmiowb();
1860
1861This is a variation on the mandatory write barrier that causes writes to weakly
1862ordered I/O regions to be partially ordered. Its effects may go beyond the
1863CPU->Hardware interface and actually affect the hardware at some level.
1864
SeongJae Park166bda72016-04-12 08:52:50 -07001865See the subsection "Acquires vs I/O accesses" for more information.
David Howells108b42b2006-03-31 16:00:29 +01001866
1867
1868===============================
1869IMPLICIT KERNEL MEMORY BARRIERS
1870===============================
1871
1872Some of the other functions in the linux kernel imply memory barriers, amongst
David Howells670bd952006-06-10 09:54:12 -07001873which are locking and scheduling functions.
David Howells108b42b2006-03-31 16:00:29 +01001874
1875This specification is a _minimum_ guarantee; any particular architecture may
1876provide more substantial guarantees, but these may not be relied upon outside
1877of arch specific code.
1878
1879
SeongJae Park166bda72016-04-12 08:52:50 -07001880LOCK ACQUISITION FUNCTIONS
1881--------------------------
David Howells108b42b2006-03-31 16:00:29 +01001882
1883The Linux kernel has a number of locking constructs:
1884
1885 (*) spin locks
1886 (*) R/W spin locks
1887 (*) mutexes
1888 (*) semaphores
1889 (*) R/W semaphores
David Howells108b42b2006-03-31 16:00:29 +01001890
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001891In all cases there are variants on "ACQUIRE" operations and "RELEASE" operations
David Howells108b42b2006-03-31 16:00:29 +01001892for each construct. These operations all imply certain barriers:
1893
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001894 (1) ACQUIRE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001895
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001896 Memory operations issued after the ACQUIRE will be completed after the
1897 ACQUIRE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001898
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001899 Memory operations issued before the ACQUIRE may be completed after
1900 the ACQUIRE operation has completed. An smp_mb__before_spinlock(),
Will Deacond9560282015-03-31 09:39:41 +01001901 combined with a following ACQUIRE, orders prior stores against
SeongJae Park0b6fa342016-04-12 08:52:53 -07001902 subsequent loads and stores. Note that this is weaker than smp_mb()!
Will Deacond9560282015-03-31 09:39:41 +01001903 The smp_mb__before_spinlock() primitive is free on many architectures.
David Howells108b42b2006-03-31 16:00:29 +01001904
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001905 (2) RELEASE operation implication:
David Howells108b42b2006-03-31 16:00:29 +01001906
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001907 Memory operations issued before the RELEASE will be completed before the
1908 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001909
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001910 Memory operations issued after the RELEASE may be completed before the
1911 RELEASE operation has completed.
David Howells108b42b2006-03-31 16:00:29 +01001912
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001913 (3) ACQUIRE vs ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001914
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001915 All ACQUIRE operations issued before another ACQUIRE operation will be
1916 completed before that ACQUIRE operation.
David Howells108b42b2006-03-31 16:00:29 +01001917
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001918 (4) ACQUIRE vs RELEASE implication:
David Howells108b42b2006-03-31 16:00:29 +01001919
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001920 All ACQUIRE operations issued before a RELEASE operation will be
1921 completed before the RELEASE operation.
David Howells108b42b2006-03-31 16:00:29 +01001922
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001923 (5) Failed conditional ACQUIRE implication:
David Howells108b42b2006-03-31 16:00:29 +01001924
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001925 Certain locking variants of the ACQUIRE operation may fail, either due to
1926 being unable to get the lock immediately, or due to receiving an unblocked
David Howells108b42b2006-03-31 16:00:29 +01001927 signal whilst asleep waiting for the lock to become available. Failed
1928 locks do not imply any sort of barrier.
1929
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001930[!] Note: one of the consequences of lock ACQUIREs and RELEASEs being only
1931one-way barriers is that the effects of instructions outside of a critical
1932section may seep into the inside of the critical section.
David Howells108b42b2006-03-31 16:00:29 +01001933
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001934An ACQUIRE followed by a RELEASE may not be assumed to be full memory barrier
1935because it is possible for an access preceding the ACQUIRE to happen after the
1936ACQUIRE, and an access following the RELEASE to happen before the RELEASE, and
1937the two accesses can themselves then cross:
David Howells670bd952006-06-10 09:54:12 -07001938
1939 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001940 ACQUIRE M
1941 RELEASE M
David Howells670bd952006-06-10 09:54:12 -07001942 *B = b;
1943
1944may occur as:
1945
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001946 ACQUIRE M, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001947
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001948When the ACQUIRE and RELEASE are a lock acquisition and release,
1949respectively, this same reordering can occur if the lock's ACQUIRE and
1950RELEASE are to the same lock variable, but only from the perspective of
1951another CPU not holding that lock. In short, a ACQUIRE followed by an
1952RELEASE may -not- be assumed to be a full memory barrier.
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001953
Paul E. McKenney12d560f2015-07-14 18:35:23 -07001954Similarly, the reverse case of a RELEASE followed by an ACQUIRE does
1955not imply a full memory barrier. Therefore, the CPU's execution of the
1956critical sections corresponding to the RELEASE and the ACQUIRE can cross,
1957so that:
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001958
1959 *A = a;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001960 RELEASE M
1961 ACQUIRE N
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001962 *B = b;
1963
1964could occur as:
1965
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01001966 ACQUIRE N, STORE *B, STORE *A, RELEASE M
Paul E. McKenney17eb88e2013-12-11 13:59:09 -08001967
Paul E. McKenney8dd853d2014-02-23 08:34:24 -08001968It might appear that this reordering could introduce a deadlock.
1969However, this cannot happen because if such a deadlock threatened,
1970the RELEASE would simply complete, thereby avoiding the deadlock.
1971
1972 Why does this work?
1973
1974 One key point is that we are only talking about the CPU doing
1975 the reordering, not the compiler. If the compiler (or, for
1976 that matter, the developer) switched the operations, deadlock
1977 -could- occur.
1978
1979 But suppose the CPU reordered the operations. In this case,
1980 the unlock precedes the lock in the assembly code. The CPU
1981 simply elected to try executing the later lock operation first.
1982 If there is a deadlock, this lock operation will simply spin (or
1983 try to sleep, but more on that later). The CPU will eventually
1984 execute the unlock operation (which preceded the lock operation
1985 in the assembly code), which will unravel the potential deadlock,
1986 allowing the lock operation to succeed.
1987
1988 But what if the lock is a sleeplock? In that case, the code will
1989 try to enter the scheduler, where it will eventually encounter
1990 a memory barrier, which will force the earlier unlock operation
1991 to complete, again unraveling the deadlock. There might be
1992 a sleep-unlock race, but the locking primitive needs to resolve
1993 such races properly in any case.
1994
David Howells108b42b2006-03-31 16:00:29 +01001995Locks and semaphores may not provide any guarantee of ordering on UP compiled
1996systems, and so cannot be counted on in such a situation to actually achieve
1997anything at all - especially with respect to I/O accesses - unless combined
1998with interrupt disabling operations.
1999
2000See also the section on "Inter-CPU locking barrier effects".
2001
2002
2003As an example, consider the following:
2004
2005 *A = a;
2006 *B = b;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002007 ACQUIRE
David Howells108b42b2006-03-31 16:00:29 +01002008 *C = c;
2009 *D = d;
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002010 RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002011 *E = e;
2012 *F = f;
2013
2014The following sequence of events is acceptable:
2015
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002016 ACQUIRE, {*F,*A}, *E, {*C,*D}, *B, RELEASE
David Howells108b42b2006-03-31 16:00:29 +01002017
2018 [+] Note that {*F,*A} indicates a combined access.
2019
2020But none of the following are:
2021
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002022 {*F,*A}, *B, ACQUIRE, *C, *D, RELEASE, *E
2023 *A, *B, *C, ACQUIRE, *D, RELEASE, *E, *F
2024 *A, *B, ACQUIRE, *C, RELEASE, *D, *E, *F
2025 *B, ACQUIRE, *C, *D, RELEASE, {*F,*A}, *E
David Howells108b42b2006-03-31 16:00:29 +01002026
2027
2028
2029INTERRUPT DISABLING FUNCTIONS
2030-----------------------------
2031
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002032Functions that disable interrupts (ACQUIRE equivalent) and enable interrupts
2033(RELEASE equivalent) will act as compiler barriers only. So if memory or I/O
David Howells108b42b2006-03-31 16:00:29 +01002034barriers are required in such a situation, they must be provided from some
2035other means.
2036
2037
David Howells50fa6102009-04-28 15:01:38 +01002038SLEEP AND WAKE-UP FUNCTIONS
2039---------------------------
2040
2041Sleeping and waking on an event flagged in global data can be viewed as an
2042interaction between two pieces of data: the task state of the task waiting for
2043the event and the global data used to indicate the event. To make sure that
2044these appear to happen in the right order, the primitives to begin the process
2045of going to sleep, and the primitives to initiate a wake up imply certain
2046barriers.
2047
2048Firstly, the sleeper normally follows something like this sequence of events:
2049
2050 for (;;) {
2051 set_current_state(TASK_UNINTERRUPTIBLE);
2052 if (event_indicated)
2053 break;
2054 schedule();
2055 }
2056
2057A general memory barrier is interpolated automatically by set_current_state()
2058after it has altered the task state:
2059
2060 CPU 1
2061 ===============================
2062 set_current_state();
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002063 smp_store_mb();
David Howells50fa6102009-04-28 15:01:38 +01002064 STORE current->state
2065 <general barrier>
2066 LOAD event_indicated
2067
2068set_current_state() may be wrapped by:
2069
2070 prepare_to_wait();
2071 prepare_to_wait_exclusive();
2072
2073which therefore also imply a general memory barrier after setting the state.
2074The whole sequence above is available in various canned forms, all of which
2075interpolate the memory barrier in the right place:
2076
2077 wait_event();
2078 wait_event_interruptible();
2079 wait_event_interruptible_exclusive();
2080 wait_event_interruptible_timeout();
2081 wait_event_killable();
2082 wait_event_timeout();
2083 wait_on_bit();
2084 wait_on_bit_lock();
2085
2086
2087Secondly, code that performs a wake up normally follows something like this:
2088
2089 event_indicated = 1;
2090 wake_up(&event_wait_queue);
2091
2092or:
2093
2094 event_indicated = 1;
2095 wake_up_process(event_daemon);
2096
SeongJae Park0b6fa342016-04-12 08:52:53 -07002097A write memory barrier is implied by wake_up() and co. if and only if they
2098wake something up. The barrier occurs before the task state is cleared, and so
2099sits between the STORE to indicate the event and the STORE to set TASK_RUNNING:
David Howells50fa6102009-04-28 15:01:38 +01002100
2101 CPU 1 CPU 2
2102 =============================== ===============================
2103 set_current_state(); STORE event_indicated
Peter Zijlstrab92b8b32015-05-12 10:51:55 +02002104 smp_store_mb(); wake_up();
David Howells50fa6102009-04-28 15:01:38 +01002105 STORE current->state <write barrier>
2106 <general barrier> STORE current->state
2107 LOAD event_indicated
2108
Paul E. McKenney5726ce02014-05-13 10:14:51 -07002109To repeat, this write memory barrier is present if and only if something
2110is actually awakened. To see this, consider the following sequence of
2111events, where X and Y are both initially zero:
2112
2113 CPU 1 CPU 2
2114 =============================== ===============================
2115 X = 1; STORE event_indicated
2116 smp_mb(); wake_up();
2117 Y = 1; wait_event(wq, Y == 1);
2118 wake_up(); load from Y sees 1, no memory barrier
2119 load from X might see 0
2120
2121In contrast, if a wakeup does occur, CPU 2's load from X would be guaranteed
2122to see 1.
2123
David Howells50fa6102009-04-28 15:01:38 +01002124The available waker functions include:
2125
2126 complete();
2127 wake_up();
2128 wake_up_all();
2129 wake_up_bit();
2130 wake_up_interruptible();
2131 wake_up_interruptible_all();
2132 wake_up_interruptible_nr();
2133 wake_up_interruptible_poll();
2134 wake_up_interruptible_sync();
2135 wake_up_interruptible_sync_poll();
2136 wake_up_locked();
2137 wake_up_locked_poll();
2138 wake_up_nr();
2139 wake_up_poll();
2140 wake_up_process();
2141
2142
2143[!] Note that the memory barriers implied by the sleeper and the waker do _not_
2144order multiple stores before the wake-up with respect to loads of those stored
2145values after the sleeper has called set_current_state(). For instance, if the
2146sleeper does:
2147
2148 set_current_state(TASK_INTERRUPTIBLE);
2149 if (event_indicated)
2150 break;
2151 __set_current_state(TASK_RUNNING);
2152 do_something(my_data);
2153
2154and the waker does:
2155
2156 my_data = value;
2157 event_indicated = 1;
2158 wake_up(&event_wait_queue);
2159
2160there's no guarantee that the change to event_indicated will be perceived by
2161the sleeper as coming after the change to my_data. In such a circumstance, the
2162code on both sides must interpolate its own memory barriers between the
2163separate data accesses. Thus the above sleeper ought to do:
2164
2165 set_current_state(TASK_INTERRUPTIBLE);
2166 if (event_indicated) {
2167 smp_rmb();
2168 do_something(my_data);
2169 }
2170
2171and the waker should do:
2172
2173 my_data = value;
2174 smp_wmb();
2175 event_indicated = 1;
2176 wake_up(&event_wait_queue);
2177
2178
David Howells108b42b2006-03-31 16:00:29 +01002179MISCELLANEOUS FUNCTIONS
2180-----------------------
2181
2182Other functions that imply barriers:
2183
2184 (*) schedule() and similar imply full memory barriers.
2185
David Howells108b42b2006-03-31 16:00:29 +01002186
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002187===================================
2188INTER-CPU ACQUIRING BARRIER EFFECTS
2189===================================
David Howells108b42b2006-03-31 16:00:29 +01002190
2191On SMP systems locking primitives give a more substantial form of barrier: one
2192that does affect memory access ordering on other CPUs, within the context of
2193conflict on any particular lock.
2194
2195
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002196ACQUIRES VS MEMORY ACCESSES
2197---------------------------
David Howells108b42b2006-03-31 16:00:29 +01002198
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002199Consider the following: the system has a pair of spinlocks (M) and (Q), and
David Howells108b42b2006-03-31 16:00:29 +01002200three CPUs; then should the following sequence of events occur:
2201
2202 CPU 1 CPU 2
2203 =============================== ===============================
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002204 WRITE_ONCE(*A, a); WRITE_ONCE(*E, e);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002205 ACQUIRE M ACQUIRE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002206 WRITE_ONCE(*B, b); WRITE_ONCE(*F, f);
2207 WRITE_ONCE(*C, c); WRITE_ONCE(*G, g);
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002208 RELEASE M RELEASE Q
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002209 WRITE_ONCE(*D, d); WRITE_ONCE(*H, h);
David Howells108b42b2006-03-31 16:00:29 +01002210
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002211Then there is no guarantee as to what order CPU 3 will see the accesses to *A
David Howells108b42b2006-03-31 16:00:29 +01002212through *H occur in, other than the constraints imposed by the separate locks
SeongJae Park0b6fa342016-04-12 08:52:53 -07002213on the separate CPUs. It might, for example, see:
David Howells108b42b2006-03-31 16:00:29 +01002214
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002215 *E, ACQUIRE M, ACQUIRE Q, *G, *C, *F, *A, *B, RELEASE Q, *D, *H, RELEASE M
David Howells108b42b2006-03-31 16:00:29 +01002216
2217But it won't see any of:
2218
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002219 *B, *C or *D preceding ACQUIRE M
2220 *A, *B or *C following RELEASE M
2221 *F, *G or *H preceding ACQUIRE Q
2222 *E, *F or *G following RELEASE Q
David Howells108b42b2006-03-31 16:00:29 +01002223
2224
David Howells108b42b2006-03-31 16:00:29 +01002225
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002226ACQUIRES VS I/O ACCESSES
2227------------------------
David Howells108b42b2006-03-31 16:00:29 +01002228
2229Under certain circumstances (especially involving NUMA), I/O accesses within
2230two spinlocked sections on two different CPUs may be seen as interleaved by the
2231PCI bridge, because the PCI bridge does not necessarily participate in the
2232cache-coherence protocol, and is therefore incapable of issuing the required
2233read memory barriers.
2234
2235For example:
2236
2237 CPU 1 CPU 2
2238 =============================== ===============================
2239 spin_lock(Q)
2240 writel(0, ADDR)
2241 writel(1, DATA);
2242 spin_unlock(Q);
2243 spin_lock(Q);
2244 writel(4, ADDR);
2245 writel(5, DATA);
2246 spin_unlock(Q);
2247
2248may be seen by the PCI bridge as follows:
2249
2250 STORE *ADDR = 0, STORE *ADDR = 4, STORE *DATA = 1, STORE *DATA = 5
2251
2252which would probably cause the hardware to malfunction.
2253
2254
2255What is necessary here is to intervene with an mmiowb() before dropping the
2256spinlock, for example:
2257
2258 CPU 1 CPU 2
2259 =============================== ===============================
2260 spin_lock(Q)
2261 writel(0, ADDR)
2262 writel(1, DATA);
2263 mmiowb();
2264 spin_unlock(Q);
2265 spin_lock(Q);
2266 writel(4, ADDR);
2267 writel(5, DATA);
2268 mmiowb();
2269 spin_unlock(Q);
2270
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002271this will ensure that the two stores issued on CPU 1 appear at the PCI bridge
2272before either of the stores issued on CPU 2.
David Howells108b42b2006-03-31 16:00:29 +01002273
2274
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002275Furthermore, following a store by a load from the same device obviates the need
2276for the mmiowb(), because the load forces the store to complete before the load
David Howells108b42b2006-03-31 16:00:29 +01002277is performed:
2278
2279 CPU 1 CPU 2
2280 =============================== ===============================
2281 spin_lock(Q)
2282 writel(0, ADDR)
2283 a = readl(DATA);
2284 spin_unlock(Q);
2285 spin_lock(Q);
2286 writel(4, ADDR);
2287 b = readl(DATA);
2288 spin_unlock(Q);
2289
2290
2291See Documentation/DocBook/deviceiobook.tmpl for more information.
2292
2293
2294=================================
2295WHERE ARE MEMORY BARRIERS NEEDED?
2296=================================
2297
2298Under normal operation, memory operation reordering is generally not going to
2299be a problem as a single-threaded linear piece of code will still appear to
David Howells50fa6102009-04-28 15:01:38 +01002300work correctly, even if it's in an SMP kernel. There are, however, four
David Howells108b42b2006-03-31 16:00:29 +01002301circumstances in which reordering definitely _could_ be a problem:
2302
2303 (*) Interprocessor interaction.
2304
2305 (*) Atomic operations.
2306
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002307 (*) Accessing devices.
David Howells108b42b2006-03-31 16:00:29 +01002308
2309 (*) Interrupts.
2310
2311
2312INTERPROCESSOR INTERACTION
2313--------------------------
2314
2315When there's a system with more than one processor, more than one CPU in the
2316system may be working on the same data set at the same time. This can cause
2317synchronisation problems, and the usual way of dealing with them is to use
2318locks. Locks, however, are quite expensive, and so it may be preferable to
2319operate without the use of a lock if at all possible. In such a case
2320operations that affect both CPUs may have to be carefully ordered to prevent
2321a malfunction.
2322
2323Consider, for example, the R/W semaphore slow path. Here a waiting process is
2324queued on the semaphore, by virtue of it having a piece of its stack linked to
2325the semaphore's list of waiting processes:
2326
2327 struct rw_semaphore {
2328 ...
2329 spinlock_t lock;
2330 struct list_head waiters;
2331 };
2332
2333 struct rwsem_waiter {
2334 struct list_head list;
2335 struct task_struct *task;
2336 };
2337
2338To wake up a particular waiter, the up_read() or up_write() functions have to:
2339
2340 (1) read the next pointer from this waiter's record to know as to where the
2341 next waiter record is;
2342
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002343 (2) read the pointer to the waiter's task structure;
David Howells108b42b2006-03-31 16:00:29 +01002344
2345 (3) clear the task pointer to tell the waiter it has been given the semaphore;
2346
2347 (4) call wake_up_process() on the task; and
2348
2349 (5) release the reference held on the waiter's task struct.
2350
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002351In other words, it has to perform this sequence of events:
David Howells108b42b2006-03-31 16:00:29 +01002352
2353 LOAD waiter->list.next;
2354 LOAD waiter->task;
2355 STORE waiter->task;
2356 CALL wakeup
2357 RELEASE task
2358
2359and if any of these steps occur out of order, then the whole thing may
2360malfunction.
2361
2362Once it has queued itself and dropped the semaphore lock, the waiter does not
2363get the lock again; it instead just waits for its task pointer to be cleared
2364before proceeding. Since the record is on the waiter's stack, this means that
2365if the task pointer is cleared _before_ the next pointer in the list is read,
2366another CPU might start processing the waiter and might clobber the waiter's
2367stack before the up*() function has a chance to read the next pointer.
2368
2369Consider then what might happen to the above sequence of events:
2370
2371 CPU 1 CPU 2
2372 =============================== ===============================
2373 down_xxx()
2374 Queue waiter
2375 Sleep
2376 up_yyy()
2377 LOAD waiter->task;
2378 STORE waiter->task;
2379 Woken up by other event
2380 <preempt>
2381 Resume processing
2382 down_xxx() returns
2383 call foo()
2384 foo() clobbers *waiter
2385 </preempt>
2386 LOAD waiter->list.next;
2387 --- OOPS ---
2388
2389This could be dealt with using the semaphore lock, but then the down_xxx()
2390function has to needlessly get the spinlock again after being woken up.
2391
2392The way to deal with this is to insert a general SMP memory barrier:
2393
2394 LOAD waiter->list.next;
2395 LOAD waiter->task;
2396 smp_mb();
2397 STORE waiter->task;
2398 CALL wakeup
2399 RELEASE task
2400
2401In this case, the barrier makes a guarantee that all memory accesses before the
2402barrier will appear to happen before all the memory accesses after the barrier
2403with respect to the other CPUs on the system. It does _not_ guarantee that all
2404the memory accesses before the barrier will be complete by the time the barrier
2405instruction itself is complete.
2406
2407On a UP system - where this wouldn't be a problem - the smp_mb() is just a
2408compiler barrier, thus making sure the compiler emits the instructions in the
David Howells6bc39272006-06-25 05:49:22 -07002409right order without actually intervening in the CPU. Since there's only one
2410CPU, that CPU's dependency ordering logic will take care of everything else.
David Howells108b42b2006-03-31 16:00:29 +01002411
2412
2413ATOMIC OPERATIONS
2414-----------------
2415
David Howellsdbc87002006-04-10 22:54:23 -07002416Whilst they are technically interprocessor interaction considerations, atomic
2417operations are noted specially as some of them imply full memory barriers and
2418some don't, but they're very heavily relied on as a group throughout the
2419kernel.
2420
2421Any atomic operation that modifies some state in memory and returns information
2422about the state (old or new) implies an SMP-conditional general memory barrier
Nick Piggin26333572007-10-18 03:06:39 -07002423(smp_mb()) on each side of the actual operation (with the exception of
2424explicit lock operations, described later). These include:
David Howells108b42b2006-03-31 16:00:29 +01002425
2426 xchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002427 atomic_xchg(); atomic_long_xchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002428 atomic_inc_return(); atomic_long_inc_return();
2429 atomic_dec_return(); atomic_long_dec_return();
2430 atomic_add_return(); atomic_long_add_return();
2431 atomic_sub_return(); atomic_long_sub_return();
2432 atomic_inc_and_test(); atomic_long_inc_and_test();
2433 atomic_dec_and_test(); atomic_long_dec_and_test();
2434 atomic_sub_and_test(); atomic_long_sub_and_test();
2435 atomic_add_negative(); atomic_long_add_negative();
David Howellsdbc87002006-04-10 22:54:23 -07002436 test_and_set_bit();
2437 test_and_clear_bit();
2438 test_and_change_bit();
David Howells108b42b2006-03-31 16:00:29 +01002439
Will Deaconed2de9f2015-07-16 16:10:06 +01002440 /* when succeeds */
2441 cmpxchg();
2442 atomic_cmpxchg(); atomic_long_cmpxchg();
Paul E. McKenneyfb2b5812013-12-11 13:59:05 -08002443 atomic_add_unless(); atomic_long_add_unless();
2444
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002445These are used for such things as implementing ACQUIRE-class and RELEASE-class
David Howellsdbc87002006-04-10 22:54:23 -07002446operations and adjusting reference counters towards object destruction, and as
2447such the implicit memory barrier effects are necessary.
David Howells108b42b2006-03-31 16:00:29 +01002448
David Howells108b42b2006-03-31 16:00:29 +01002449
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002450The following operations are potential problems as they do _not_ imply memory
Peter Zijlstra2e4f5382013-11-06 14:57:36 +01002451barriers, but might be used for implementing such things as RELEASE-class
David Howellsdbc87002006-04-10 22:54:23 -07002452operations:
2453
2454 atomic_set();
David Howells108b42b2006-03-31 16:00:29 +01002455 set_bit();
2456 clear_bit();
2457 change_bit();
David Howellsdbc87002006-04-10 22:54:23 -07002458
2459With these the appropriate explicit memory barrier should be used if necessary
Peter Zijlstra1b156112014-03-13 19:00:35 +01002460(smp_mb__before_atomic() for instance).
David Howells108b42b2006-03-31 16:00:29 +01002461
2462
David Howellsdbc87002006-04-10 22:54:23 -07002463The following also do _not_ imply memory barriers, and so may require explicit
Peter Zijlstra1b156112014-03-13 19:00:35 +01002464memory barriers under some circumstances (smp_mb__before_atomic() for
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002465instance):
David Howells108b42b2006-03-31 16:00:29 +01002466
2467 atomic_add();
2468 atomic_sub();
2469 atomic_inc();
2470 atomic_dec();
2471
2472If they're used for statistics generation, then they probably don't need memory
2473barriers, unless there's a coupling between statistical data.
2474
2475If they're used for reference counting on an object to control its lifetime,
2476they probably don't need memory barriers because either the reference count
2477will be adjusted inside a locked section, or the caller will already hold
2478sufficient references to make the lock, and thus a memory barrier unnecessary.
2479
2480If they're used for constructing a lock of some description, then they probably
2481do need memory barriers as a lock primitive generally has to do things in a
2482specific order.
2483
David Howells108b42b2006-03-31 16:00:29 +01002484Basically, each usage case has to be carefully considered as to whether memory
David Howellsdbc87002006-04-10 22:54:23 -07002485barriers are needed or not.
2486
Nick Piggin26333572007-10-18 03:06:39 -07002487The following operations are special locking primitives:
2488
2489 test_and_set_bit_lock();
2490 clear_bit_unlock();
2491 __clear_bit_unlock();
2492
SeongJae Park0b6fa342016-04-12 08:52:53 -07002493These implement ACQUIRE-class and RELEASE-class operations. These should be
2494used in preference to other operations when implementing locking primitives,
2495because their implementations can be optimised on many architectures.
Nick Piggin26333572007-10-18 03:06:39 -07002496
David Howellsdbc87002006-04-10 22:54:23 -07002497[!] Note that special memory barrier primitives are available for these
2498situations because on some CPUs the atomic instructions used imply full memory
2499barriers, and so barrier instructions are superfluous in conjunction with them,
2500and in such cases the special barrier primitives will be no-ops.
David Howells108b42b2006-03-31 16:00:29 +01002501
2502See Documentation/atomic_ops.txt for more information.
2503
2504
2505ACCESSING DEVICES
2506-----------------
2507
2508Many devices can be memory mapped, and so appear to the CPU as if they're just
2509a set of memory locations. To control such a device, the driver usually has to
2510make the right memory accesses in exactly the right order.
2511
2512However, having a clever CPU or a clever compiler creates a potential problem
2513in that the carefully sequenced accesses in the driver code won't reach the
2514device in the requisite order if the CPU or the compiler thinks it is more
2515efficient to reorder, combine or merge accesses - something that would cause
2516the device to malfunction.
2517
2518Inside of the Linux kernel, I/O should be done through the appropriate accessor
2519routines - such as inb() or writel() - which know how to make such accesses
2520appropriately sequential. Whilst this, for the most part, renders the explicit
2521use of memory barriers unnecessary, there are a couple of situations where they
2522might be needed:
2523
2524 (1) On some systems, I/O stores are not strongly ordered across all CPUs, and
2525 so for _all_ general drivers locks should be used and mmiowb() must be
2526 issued prior to unlocking the critical section.
2527
2528 (2) If the accessor functions are used to refer to an I/O memory window with
2529 relaxed memory access properties, then _mandatory_ memory barriers are
2530 required to enforce ordering.
2531
2532See Documentation/DocBook/deviceiobook.tmpl for more information.
2533
2534
2535INTERRUPTS
2536----------
2537
2538A driver may be interrupted by its own interrupt service routine, and thus the
2539two parts of the driver may interfere with each other's attempts to control or
2540access the device.
2541
2542This may be alleviated - at least in part - by disabling local interrupts (a
2543form of locking), such that the critical operations are all contained within
2544the interrupt-disabled section in the driver. Whilst the driver's interrupt
2545routine is executing, the driver's core may not run on the same CPU, and its
2546interrupt is not permitted to happen again until the current interrupt has been
2547handled, thus the interrupt handler does not need to lock against that.
2548
2549However, consider a driver that was talking to an ethernet card that sports an
2550address register and a data register. If that driver's core talks to the card
2551under interrupt-disablement and then the driver's interrupt handler is invoked:
2552
2553 LOCAL IRQ DISABLE
2554 writew(ADDR, 3);
2555 writew(DATA, y);
2556 LOCAL IRQ ENABLE
2557 <interrupt>
2558 writew(ADDR, 4);
2559 q = readw(DATA);
2560 </interrupt>
2561
2562The store to the data register might happen after the second store to the
2563address register if ordering rules are sufficiently relaxed:
2564
2565 STORE *ADDR = 3, STORE *ADDR = 4, STORE *DATA = y, q = LOAD *DATA
2566
2567
2568If ordering rules are relaxed, it must be assumed that accesses done inside an
2569interrupt disabled section may leak outside of it and may interleave with
2570accesses performed in an interrupt - and vice versa - unless implicit or
2571explicit barriers are used.
2572
2573Normally this won't be a problem because the I/O accesses done inside such
2574sections will include synchronous load operations on strictly ordered I/O
SeongJae Park0b6fa342016-04-12 08:52:53 -07002575registers that form implicit I/O barriers. If this isn't sufficient then an
David Howells108b42b2006-03-31 16:00:29 +01002576mmiowb() may need to be used explicitly.
2577
2578
2579A similar situation may occur between an interrupt routine and two routines
SeongJae Park0b6fa342016-04-12 08:52:53 -07002580running on separate CPUs that communicate with each other. If such a case is
David Howells108b42b2006-03-31 16:00:29 +01002581likely, then interrupt-disabling locks should be used to guarantee ordering.
2582
2583
2584==========================
2585KERNEL I/O BARRIER EFFECTS
2586==========================
2587
2588When accessing I/O memory, drivers should use the appropriate accessor
2589functions:
2590
2591 (*) inX(), outX():
2592
2593 These are intended to talk to I/O space rather than memory space, but
SeongJae Park0b6fa342016-04-12 08:52:53 -07002594 that's primarily a CPU-specific concept. The i386 and x86_64 processors
2595 do indeed have special I/O space access cycles and instructions, but many
David Howells108b42b2006-03-31 16:00:29 +01002596 CPUs don't have such a concept.
2597
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002598 The PCI bus, amongst others, defines an I/O space concept which - on such
2599 CPUs as i386 and x86_64 - readily maps to the CPU's concept of I/O
David Howells6bc39272006-06-25 05:49:22 -07002600 space. However, it may also be mapped as a virtual I/O space in the CPU's
2601 memory map, particularly on those CPUs that don't support alternate I/O
2602 spaces.
David Howells108b42b2006-03-31 16:00:29 +01002603
2604 Accesses to this space may be fully synchronous (as on i386), but
2605 intermediary bridges (such as the PCI host bridge) may not fully honour
2606 that.
2607
2608 They are guaranteed to be fully ordered with respect to each other.
2609
2610 They are not guaranteed to be fully ordered with respect to other types of
2611 memory and I/O operation.
2612
2613 (*) readX(), writeX():
2614
2615 Whether these are guaranteed to be fully ordered and uncombined with
2616 respect to each other on the issuing CPU depends on the characteristics
SeongJae Park0b6fa342016-04-12 08:52:53 -07002617 defined for the memory window through which they're accessing. On later
David Howells108b42b2006-03-31 16:00:29 +01002618 i386 architecture machines, for example, this is controlled by way of the
2619 MTRR registers.
2620
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002621 Ordinarily, these will be guaranteed to be fully ordered and uncombined,
David Howells108b42b2006-03-31 16:00:29 +01002622 provided they're not accessing a prefetchable device.
2623
2624 However, intermediary hardware (such as a PCI bridge) may indulge in
2625 deferral if it so wishes; to flush a store, a load from the same location
2626 is preferred[*], but a load from the same device or from configuration
2627 space should suffice for PCI.
2628
2629 [*] NOTE! attempting to load from the same location as was written to may
Ingo Molnare0edc782013-11-22 11:24:53 +01002630 cause a malfunction - consider the 16550 Rx/Tx serial registers for
2631 example.
David Howells108b42b2006-03-31 16:00:29 +01002632
2633 Used with prefetchable I/O memory, an mmiowb() barrier may be required to
2634 force stores to be ordered.
2635
2636 Please refer to the PCI specification for more information on interactions
2637 between PCI transactions.
2638
Will Deacona8e0aea2013-09-04 12:30:08 +01002639 (*) readX_relaxed(), writeX_relaxed()
David Howells108b42b2006-03-31 16:00:29 +01002640
Will Deacona8e0aea2013-09-04 12:30:08 +01002641 These are similar to readX() and writeX(), but provide weaker memory
SeongJae Park0b6fa342016-04-12 08:52:53 -07002642 ordering guarantees. Specifically, they do not guarantee ordering with
Will Deacona8e0aea2013-09-04 12:30:08 +01002643 respect to normal memory accesses (e.g. DMA buffers) nor do they guarantee
SeongJae Park0b6fa342016-04-12 08:52:53 -07002644 ordering with respect to LOCK or UNLOCK operations. If the latter is
2645 required, an mmiowb() barrier can be used. Note that relaxed accesses to
Will Deacona8e0aea2013-09-04 12:30:08 +01002646 the same peripheral are guaranteed to be ordered with respect to each
2647 other.
David Howells108b42b2006-03-31 16:00:29 +01002648
2649 (*) ioreadX(), iowriteX()
2650
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002651 These will perform appropriately for the type of access they're actually
David Howells108b42b2006-03-31 16:00:29 +01002652 doing, be it inX()/outX() or readX()/writeX().
2653
2654
2655========================================
2656ASSUMED MINIMUM EXECUTION ORDERING MODEL
2657========================================
2658
2659It has to be assumed that the conceptual CPU is weakly-ordered but that it will
2660maintain the appearance of program causality with respect to itself. Some CPUs
2661(such as i386 or x86_64) are more constrained than others (such as powerpc or
2662frv), and so the most relaxed case (namely DEC Alpha) must be assumed outside
2663of arch-specific code.
2664
2665This means that it must be considered that the CPU will execute its instruction
2666stream in any order it feels like - or even in parallel - provided that if an
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002667instruction in the stream depends on an earlier instruction, then that
David Howells108b42b2006-03-31 16:00:29 +01002668earlier instruction must be sufficiently complete[*] before the later
2669instruction may proceed; in other words: provided that the appearance of
2670causality is maintained.
2671
2672 [*] Some instructions have more than one effect - such as changing the
2673 condition codes, changing registers or changing memory - and different
2674 instructions may depend on different effects.
2675
2676A CPU may also discard any instruction sequence that winds up having no
2677ultimate effect. For example, if two adjacent instructions both load an
2678immediate value into the same register, the first may be discarded.
2679
2680
2681Similarly, it has to be assumed that compiler might reorder the instruction
2682stream in any way it sees fit, again provided the appearance of causality is
2683maintained.
2684
2685
2686============================
2687THE EFFECTS OF THE CPU CACHE
2688============================
2689
2690The way cached memory operations are perceived across the system is affected to
2691a certain extent by the caches that lie between CPUs and memory, and by the
2692memory coherence system that maintains the consistency of state in the system.
2693
2694As far as the way a CPU interacts with another part of the system through the
2695caches goes, the memory system has to include the CPU's caches, and memory
2696barriers for the most part act at the interface between the CPU and its cache
2697(memory barriers logically act on the dotted line in the following diagram):
2698
2699 <--- CPU ---> : <----------- Memory ----------->
2700 :
2701 +--------+ +--------+ : +--------+ +-----------+
2702 | | | | : | | | | +--------+
Ingo Molnare0edc782013-11-22 11:24:53 +01002703 | CPU | | Memory | : | CPU | | | | |
2704 | Core |--->| Access |----->| Cache |<-->| | | |
David Howells108b42b2006-03-31 16:00:29 +01002705 | | | Queue | : | | | |--->| Memory |
Ingo Molnare0edc782013-11-22 11:24:53 +01002706 | | | | : | | | | | |
2707 +--------+ +--------+ : +--------+ | | | |
David Howells108b42b2006-03-31 16:00:29 +01002708 : | Cache | +--------+
2709 : | Coherency |
2710 : | Mechanism | +--------+
2711 +--------+ +--------+ : +--------+ | | | |
2712 | | | | : | | | | | |
2713 | CPU | | Memory | : | CPU | | |--->| Device |
Ingo Molnare0edc782013-11-22 11:24:53 +01002714 | Core |--->| Access |----->| Cache |<-->| | | |
2715 | | | Queue | : | | | | | |
David Howells108b42b2006-03-31 16:00:29 +01002716 | | | | : | | | | +--------+
2717 +--------+ +--------+ : +--------+ +-----------+
2718 :
2719 :
2720
2721Although any particular load or store may not actually appear outside of the
2722CPU that issued it since it may have been satisfied within the CPU's own cache,
2723it will still appear as if the full memory access had taken place as far as the
2724other CPUs are concerned since the cache coherency mechanisms will migrate the
2725cacheline over to the accessing CPU and propagate the effects upon conflict.
2726
2727The CPU core may execute instructions in any order it deems fit, provided the
2728expected program causality appears to be maintained. Some of the instructions
2729generate load and store operations which then go into the queue of memory
2730accesses to be performed. The core may place these in the queue in any order
2731it wishes, and continue execution until it is forced to wait for an instruction
2732to complete.
2733
2734What memory barriers are concerned with is controlling the order in which
2735accesses cross from the CPU side of things to the memory side of things, and
2736the order in which the effects are perceived to happen by the other observers
2737in the system.
2738
2739[!] Memory barriers are _not_ needed within a given CPU, as CPUs always see
2740their own loads and stores as if they had happened in program order.
2741
2742[!] MMIO or other device accesses may bypass the cache system. This depends on
2743the properties of the memory window through which devices are accessed and/or
2744the use of any special device communication instructions the CPU may have.
2745
2746
2747CACHE COHERENCY
2748---------------
2749
2750Life isn't quite as simple as it may appear above, however: for while the
2751caches are expected to be coherent, there's no guarantee that that coherency
2752will be ordered. This means that whilst changes made on one CPU will
2753eventually become visible on all CPUs, there's no guarantee that they will
2754become apparent in the same order on those other CPUs.
2755
2756
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002757Consider dealing with a system that has a pair of CPUs (1 & 2), each of which
2758has a pair of parallel data caches (CPU 1 has A/B, and CPU 2 has C/D):
David Howells108b42b2006-03-31 16:00:29 +01002759
2760 :
2761 : +--------+
2762 : +---------+ | |
2763 +--------+ : +--->| Cache A |<------->| |
2764 | | : | +---------+ | |
2765 | CPU 1 |<---+ | |
2766 | | : | +---------+ | |
2767 +--------+ : +--->| Cache B |<------->| |
2768 : +---------+ | |
2769 : | Memory |
2770 : +---------+ | System |
2771 +--------+ : +--->| Cache C |<------->| |
2772 | | : | +---------+ | |
2773 | CPU 2 |<---+ | |
2774 | | : | +---------+ | |
2775 +--------+ : +--->| Cache D |<------->| |
2776 : +---------+ | |
2777 : +--------+
2778 :
2779
2780Imagine the system has the following properties:
2781
2782 (*) an odd-numbered cache line may be in cache A, cache C or it may still be
2783 resident in memory;
2784
2785 (*) an even-numbered cache line may be in cache B, cache D or it may still be
2786 resident in memory;
2787
2788 (*) whilst the CPU core is interrogating one cache, the other cache may be
2789 making use of the bus to access the rest of the system - perhaps to
2790 displace a dirty cacheline or to do a speculative load;
2791
2792 (*) each cache has a queue of operations that need to be applied to that cache
2793 to maintain coherency with the rest of the system;
2794
2795 (*) the coherency queue is not flushed by normal loads to lines already
2796 present in the cache, even though the contents of the queue may
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002797 potentially affect those loads.
David Howells108b42b2006-03-31 16:00:29 +01002798
2799Imagine, then, that two writes are made on the first CPU, with a write barrier
2800between them to guarantee that they will appear to reach that CPU's caches in
2801the requisite order:
2802
2803 CPU 1 CPU 2 COMMENT
2804 =============== =============== =======================================
2805 u == 0, v == 1 and p == &u, q == &u
2806 v = 2;
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002807 smp_wmb(); Make sure change to v is visible before
David Howells108b42b2006-03-31 16:00:29 +01002808 change to p
2809 <A:modify v=2> v is now in cache A exclusively
2810 p = &v;
2811 <B:modify p=&v> p is now in cache B exclusively
2812
2813The write memory barrier forces the other CPUs in the system to perceive that
2814the local CPU's caches have apparently been updated in the correct order. But
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002815now imagine that the second CPU wants to read those values:
David Howells108b42b2006-03-31 16:00:29 +01002816
2817 CPU 1 CPU 2 COMMENT
2818 =============== =============== =======================================
2819 ...
2820 q = p;
2821 x = *q;
2822
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002823The above pair of reads may then fail to happen in the expected order, as the
David Howells108b42b2006-03-31 16:00:29 +01002824cacheline holding p may get updated in one of the second CPU's caches whilst
2825the update to the cacheline holding v is delayed in the other of the second
2826CPU's caches by some other cache event:
2827
2828 CPU 1 CPU 2 COMMENT
2829 =============== =============== =======================================
2830 u == 0, v == 1 and p == &u, q == &u
2831 v = 2;
2832 smp_wmb();
2833 <A:modify v=2> <C:busy>
2834 <C:queue v=2>
Aneesh Kumar79afecf2006-05-15 09:44:36 -07002835 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002836 <D:request p>
2837 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002838 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002839 x = *q;
2840 <C:read *q> Reads from v before v updated in cache
2841 <C:unbusy>
2842 <C:commit v=2>
2843
2844Basically, whilst both cachelines will be updated on CPU 2 eventually, there's
2845no guarantee that, without intervention, the order of update will be the same
2846as that committed on CPU 1.
2847
2848
2849To intervene, we need to interpolate a data dependency barrier or a read
2850barrier between the loads. This will force the cache to commit its coherency
2851queue before processing any further requests:
2852
2853 CPU 1 CPU 2 COMMENT
2854 =============== =============== =======================================
2855 u == 0, v == 1 and p == &u, q == &u
2856 v = 2;
2857 smp_wmb();
2858 <A:modify v=2> <C:busy>
2859 <C:queue v=2>
Paolo 'Blaisorblade' Giarrusso3fda9822006-10-19 23:28:19 -07002860 p = &v; q = p;
David Howells108b42b2006-03-31 16:00:29 +01002861 <D:request p>
2862 <B:modify p=&v> <D:commit p=&v>
Ingo Molnare0edc782013-11-22 11:24:53 +01002863 <D:read p>
David Howells108b42b2006-03-31 16:00:29 +01002864 smp_read_barrier_depends()
2865 <C:unbusy>
2866 <C:commit v=2>
2867 x = *q;
2868 <C:read *q> Reads from v after v updated in cache
2869
2870
2871This sort of problem can be encountered on DEC Alpha processors as they have a
2872split cache that improves performance by making better use of the data bus.
2873Whilst most CPUs do imply a data dependency barrier on the read when a memory
2874access depends on a read, not all do, so it may not be relied on.
2875
2876Other CPUs may also have split caches, but must coordinate between the various
Matt LaPlante3f6dee92006-10-03 22:45:33 +02002877cachelets for normal memory accesses. The semantics of the Alpha removes the
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002878need for coordination in the absence of memory barriers.
David Howells108b42b2006-03-31 16:00:29 +01002879
2880
2881CACHE COHERENCY VS DMA
2882----------------------
2883
2884Not all systems maintain cache coherency with respect to devices doing DMA. In
2885such cases, a device attempting DMA may obtain stale data from RAM because
2886dirty cache lines may be resident in the caches of various CPUs, and may not
2887have been written back to RAM yet. To deal with this, the appropriate part of
2888the kernel must flush the overlapping bits of cache on each CPU (and maybe
2889invalidate them as well).
2890
2891In addition, the data DMA'd to RAM by a device may be overwritten by dirty
2892cache lines being written back to RAM from a CPU's cache after the device has
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002893installed its own data, or cache lines present in the CPU's cache may simply
2894obscure the fact that RAM has been updated, until at such time as the cacheline
2895is discarded from the CPU's cache and reloaded. To deal with this, the
2896appropriate part of the kernel must invalidate the overlapping bits of the
David Howells108b42b2006-03-31 16:00:29 +01002897cache on each CPU.
2898
2899See Documentation/cachetlb.txt for more information on cache management.
2900
2901
2902CACHE COHERENCY VS MMIO
2903-----------------------
2904
2905Memory mapped I/O usually takes place through memory locations that are part of
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002906a window in the CPU's memory space that has different properties assigned than
David Howells108b42b2006-03-31 16:00:29 +01002907the usual RAM directed window.
2908
2909Amongst these properties is usually the fact that such accesses bypass the
2910caching entirely and go directly to the device buses. This means MMIO accesses
2911may, in effect, overtake accesses to cached memory that were emitted earlier.
2912A memory barrier isn't sufficient in such a case, but rather the cache must be
2913flushed between the cached memory write and the MMIO access if the two are in
2914any way dependent.
2915
2916
2917=========================
2918THE THINGS CPUS GET UP TO
2919=========================
2920
2921A programmer might take it for granted that the CPU will perform memory
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002922operations in exactly the order specified, so that if the CPU is, for example,
David Howells108b42b2006-03-31 16:00:29 +01002923given the following piece of code to execute:
2924
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002925 a = READ_ONCE(*A);
2926 WRITE_ONCE(*B, b);
2927 c = READ_ONCE(*C);
2928 d = READ_ONCE(*D);
2929 WRITE_ONCE(*E, e);
David Howells108b42b2006-03-31 16:00:29 +01002930
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002931they would then expect that the CPU will complete the memory operation for each
David Howells108b42b2006-03-31 16:00:29 +01002932instruction before moving on to the next one, leading to a definite sequence of
2933operations as seen by external observers in the system:
2934
2935 LOAD *A, STORE *B, LOAD *C, LOAD *D, STORE *E.
2936
2937
2938Reality is, of course, much messier. With many CPUs and compilers, the above
2939assumption doesn't hold because:
2940
2941 (*) loads are more likely to need to be completed immediately to permit
2942 execution progress, whereas stores can often be deferred without a
2943 problem;
2944
2945 (*) loads may be done speculatively, and the result discarded should it prove
2946 to have been unnecessary;
2947
Jarek Poplawski81fc6322007-05-23 13:58:20 -07002948 (*) loads may be done speculatively, leading to the result having been fetched
2949 at the wrong time in the expected sequence of events;
David Howells108b42b2006-03-31 16:00:29 +01002950
2951 (*) the order of the memory accesses may be rearranged to promote better use
2952 of the CPU buses and caches;
2953
2954 (*) loads and stores may be combined to improve performance when talking to
2955 memory or I/O hardware that can do batched accesses of adjacent locations,
2956 thus cutting down on transaction setup costs (memory and PCI devices may
2957 both be able to do this); and
2958
2959 (*) the CPU's data cache may affect the ordering, and whilst cache-coherency
2960 mechanisms may alleviate this - once the store has actually hit the cache
2961 - there's no guarantee that the coherency management will be propagated in
2962 order to other CPUs.
2963
2964So what another CPU, say, might actually observe from the above piece of code
2965is:
2966
2967 LOAD *A, ..., LOAD {*C,*D}, STORE *E, STORE *B
2968
2969 (Where "LOAD {*C,*D}" is a combined load)
2970
2971
2972However, it is guaranteed that a CPU will be self-consistent: it will see its
2973_own_ accesses appear to be correctly ordered, without the need for a memory
2974barrier. For instance with the following code:
2975
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002976 U = READ_ONCE(*A);
2977 WRITE_ONCE(*A, V);
2978 WRITE_ONCE(*A, W);
2979 X = READ_ONCE(*A);
2980 WRITE_ONCE(*A, Y);
2981 Z = READ_ONCE(*A);
David Howells108b42b2006-03-31 16:00:29 +01002982
2983and assuming no intervention by an external influence, it can be assumed that
2984the final result will appear to be:
2985
2986 U == the original value of *A
2987 X == W
2988 Z == Y
2989 *A == Y
2990
2991The code above may cause the CPU to generate the full sequence of memory
2992accesses:
2993
2994 U=LOAD *A, STORE *A=V, STORE *A=W, X=LOAD *A, STORE *A=Y, Z=LOAD *A
2995
2996in that order, but, without intervention, the sequence may have almost any
Paul E. McKenney9af194c2015-06-18 14:33:24 -07002997combination of elements combined or discarded, provided the program's view
2998of the world remains consistent. Note that READ_ONCE() and WRITE_ONCE()
2999are -not- optional in the above example, as there are architectures
3000where a given CPU might reorder successive loads to the same location.
3001On such architectures, READ_ONCE() and WRITE_ONCE() do whatever is
3002necessary to prevent this, for example, on Itanium the volatile casts
3003used by READ_ONCE() and WRITE_ONCE() cause GCC to emit the special ld.acq
3004and st.rel instructions (respectively) that prevent such reordering.
David Howells108b42b2006-03-31 16:00:29 +01003005
3006The compiler may also combine, discard or defer elements of the sequence before
3007the CPU even sees them.
3008
3009For instance:
3010
3011 *A = V;
3012 *A = W;
3013
3014may be reduced to:
3015
3016 *A = W;
3017
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003018since, without either a write barrier or an WRITE_ONCE(), it can be
Paul E. McKenney2ecf8102013-12-11 13:59:04 -08003019assumed that the effect of the storage of V to *A is lost. Similarly:
David Howells108b42b2006-03-31 16:00:29 +01003020
3021 *A = Y;
3022 Z = *A;
3023
Paul E. McKenney9af194c2015-06-18 14:33:24 -07003024may, without a memory barrier or an READ_ONCE() and WRITE_ONCE(), be
3025reduced to:
David Howells108b42b2006-03-31 16:00:29 +01003026
3027 *A = Y;
3028 Z = Y;
3029
3030and the LOAD operation never appear outside of the CPU.
3031
3032
3033AND THEN THERE'S THE ALPHA
3034--------------------------
3035
3036The DEC Alpha CPU is one of the most relaxed CPUs there is. Not only that,
3037some versions of the Alpha CPU have a split data cache, permitting them to have
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003038two semantically-related cache lines updated at separate times. This is where
David Howells108b42b2006-03-31 16:00:29 +01003039the data dependency barrier really becomes necessary as this synchronises both
3040caches with the memory coherence system, thus making it seem like pointer
3041changes vs new data occur in the right order.
3042
Jarek Poplawski81fc6322007-05-23 13:58:20 -07003043The Alpha defines the Linux kernel's memory barrier model.
David Howells108b42b2006-03-31 16:00:29 +01003044
3045See the subsection on "Cache Coherency" above.
3046
SeongJae Park0b6fa342016-04-12 08:52:53 -07003047
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003048VIRTUAL MACHINE GUESTS
SeongJae Park3dbf0912016-04-12 08:52:52 -07003049----------------------
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003050
3051Guests running within virtual machines might be affected by SMP effects even if
3052the guest itself is compiled without SMP support. This is an artifact of
3053interfacing with an SMP host while running an UP kernel. Using mandatory
3054barriers for this use-case would be possible but is often suboptimal.
3055
3056To handle this case optimally, low-level virt_mb() etc macros are available.
3057These have the same effect as smp_mb() etc when SMP is enabled, but generate
SeongJae Park0b6fa342016-04-12 08:52:53 -07003058identical code for SMP and non-SMP systems. For example, virtual machine guests
Michael S. Tsirkin6a65d262015-12-27 18:23:01 +02003059should use virt_mb() rather than smp_mb() when synchronizing against a
3060(possibly SMP) host.
3061
3062These are equivalent to smp_mb() etc counterparts in all other respects,
3063in particular, they do not control MMIO effects: to control
3064MMIO effects, use mandatory barriers.
David Howells108b42b2006-03-31 16:00:29 +01003065
SeongJae Park0b6fa342016-04-12 08:52:53 -07003066
David Howells90fddab2010-03-24 09:43:00 +00003067============
3068EXAMPLE USES
3069============
3070
3071CIRCULAR BUFFERS
3072----------------
3073
3074Memory barriers can be used to implement circular buffering without the need
3075of a lock to serialise the producer with the consumer. See:
3076
3077 Documentation/circular-buffers.txt
3078
3079for details.
3080
3081
David Howells108b42b2006-03-31 16:00:29 +01003082==========
3083REFERENCES
3084==========
3085
3086Alpha AXP Architecture Reference Manual, Second Edition (Sites & Witek,
3087Digital Press)
3088 Chapter 5.2: Physical Address Space Characteristics
3089 Chapter 5.4: Caches and Write Buffers
3090 Chapter 5.5: Data Sharing
3091 Chapter 5.6: Read/Write Ordering
3092
3093AMD64 Architecture Programmer's Manual Volume 2: System Programming
3094 Chapter 7.1: Memory-Access Ordering
3095 Chapter 7.4: Buffering and Combining Memory Writes
3096
3097IA-32 Intel Architecture Software Developer's Manual, Volume 3:
3098System Programming Guide
3099 Chapter 7.1: Locked Atomic Operations
3100 Chapter 7.2: Memory Ordering
3101 Chapter 7.4: Serializing Instructions
3102
3103The SPARC Architecture Manual, Version 9
3104 Chapter 8: Memory Models
3105 Appendix D: Formal Specification of the Memory Models
3106 Appendix J: Programming with the Memory Models
3107
3108UltraSPARC Programmer Reference Manual
3109 Chapter 5: Memory Accesses and Cacheability
3110 Chapter 15: Sparc-V9 Memory Models
3111
3112UltraSPARC III Cu User's Manual
3113 Chapter 9: Memory Models
3114
3115UltraSPARC IIIi Processor User's Manual
3116 Chapter 8: Memory Models
3117
3118UltraSPARC Architecture 2005
3119 Chapter 9: Memory
3120 Appendix D: Formal Specifications of the Memory Models
3121
3122UltraSPARC T1 Supplement to the UltraSPARC Architecture 2005
3123 Chapter 8: Memory Models
3124 Appendix F: Caches and Cache Coherency
3125
3126Solaris Internals, Core Kernel Architecture, p63-68:
3127 Chapter 3.3: Hardware Considerations for Locks and
3128 Synchronization
3129
3130Unix Systems for Modern Architectures, Symmetric Multiprocessing and Caching
3131for Kernel Programmers:
3132 Chapter 13: Other Memory Models
3133
3134Intel Itanium Architecture Software Developer's Manual: Volume 1:
3135 Section 2.6: Speculation
3136 Section 4.4: Memory Access