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-rw-r--r--Documentation/memory-barriers.txt139
1 files changed, 100 insertions, 39 deletions
diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index 102dc19c4119..556f951f8626 100644
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -608,26 +608,30 @@ as follows:
b = p; /* BUG: Compiler can reorder!!! */
do_something();
-The solution is again ACCESS_ONCE(), which preserves the ordering between
-the load from variable 'a' and the store to variable 'b':
+The solution is again ACCESS_ONCE() and barrier(), which preserves the
+ordering between the load from variable 'a' and the store to variable 'b':
q = ACCESS_ONCE(a);
if (q) {
+ barrier();
ACCESS_ONCE(b) = p;
do_something();
} else {
+ barrier();
ACCESS_ONCE(b) = p;
do_something_else();
}
-You could also use barrier() to prevent the compiler from moving
-the stores to variable 'b', but barrier() would not prevent the
-compiler from proving to itself that a==1 always, so ACCESS_ONCE()
-is also needed.
+The initial ACCESS_ONCE() is required to prevent the compiler from
+proving the value of 'a', and the pair of barrier() invocations are
+required to prevent the compiler from pulling the two identical stores
+to 'b' out from the legs of the "if" statement.
It is important to note that control dependencies absolutely require a
a conditional. For example, the following "optimized" version of
-the above example breaks ordering:
+the above example breaks ordering, which is why the barrier() invocations
+are absolutely required if you have identical stores in both legs of
+the "if" statement:
q = ACCESS_ONCE(a);
ACCESS_ONCE(b) = p; /* BUG: No ordering vs. load from a!!! */
@@ -643,9 +647,11 @@ It is of course legal for the prior load to be part of the conditional,
for example, as follows:
if (ACCESS_ONCE(a) > 0) {
+ barrier();
ACCESS_ONCE(b) = q / 2;
do_something();
} else {
+ barrier();
ACCESS_ONCE(b) = q / 3;
do_something_else();
}
@@ -659,9 +665,11 @@ the needed conditional. For example:
q = ACCESS_ONCE(a);
if (q % MAX) {
+ barrier();
ACCESS_ONCE(b) = p;
do_something();
} else {
+ barrier();
ACCESS_ONCE(b) = p;
do_something_else();
}
@@ -723,8 +731,13 @@ In summary:
use smb_rmb(), smp_wmb(), or, in the case of prior stores and
later loads, smp_mb().
+ (*) If both legs of the "if" statement begin with identical stores
+ to the same variable, a barrier() statement is required at the
+ beginning of each leg of the "if" statement.
+
(*) Control dependencies require at least one run-time conditional
- between the prior load and the subsequent store. If the compiler
+ between the prior load and the subsequent store, and this
+ conditional must involve the prior load. If the compiler
is able to optimize the conditional away, it will have also
optimized away the ordering. Careful use of ACCESS_ONCE() can
help to preserve the needed conditional.
@@ -1249,6 +1262,23 @@ The ACCESS_ONCE() function can prevent any number of optimizations that,
while perfectly safe in single-threaded code, can be fatal in concurrent
code. Here are some examples of these sorts of optimizations:
+ (*) The compiler is within its rights to reorder loads and stores
+ to the same variable, and in some cases, the CPU is within its
+ rights to reorder loads to the same variable. This means that
+ the following code:
+
+ a[0] = x;
+ a[1] = x;
+
+ Might result in an older value of x stored in a[1] than in a[0].
+ Prevent both the compiler and the CPU from doing this as follows:
+
+ a[0] = ACCESS_ONCE(x);
+ a[1] = ACCESS_ONCE(x);
+
+ In short, ACCESS_ONCE() provides cache coherence for accesses from
+ multiple CPUs to a single variable.
+
(*) The compiler is within its rights to merge successive loads from
the same variable. Such merging can cause the compiler to "optimize"
the following code:
@@ -1371,7 +1401,7 @@ code. Here are some examples of these sorts of optimizations:
process_message(msg);
}
- There is nothing to prevent the the compiler from transforming
+ There is nothing to prevent the compiler from transforming
process_level() to the following, in fact, this might well be a
win for single-threaded code:
@@ -1644,12 +1674,12 @@ for each construct. These operations all imply certain barriers:
Memory operations issued after the ACQUIRE will be completed after the
ACQUIRE operation has completed.
- Memory operations issued before the ACQUIRE may be completed after the
- ACQUIRE operation has completed. An smp_mb__before_spinlock(), combined
- with a following ACQUIRE, orders prior loads against subsequent stores and
- stores and prior stores against subsequent stores. Note that this is
- weaker than smp_mb()! The smp_mb__before_spinlock() primitive is free on
- many architectures.
+ Memory operations issued before the ACQUIRE may be completed after
+ the ACQUIRE operation has completed. An smp_mb__before_spinlock(),
+ combined with a following ACQUIRE, orders prior loads against
+ subsequent loads and stores and also orders prior stores against
+ subsequent stores. Note that this is weaker than smp_mb()! The
+ smp_mb__before_spinlock() primitive is free on many architectures.
(2) RELEASE operation implication:
@@ -1694,24 +1724,21 @@ may occur as:
ACQUIRE M, STORE *B, STORE *A, RELEASE M
-This same reordering can of course occur if the lock's ACQUIRE and RELEASE are
-to the same lock variable, but only from the perspective of another CPU not
-holding that lock.
-
-In short, a RELEASE followed by an ACQUIRE may -not- be assumed to be a full
-memory barrier because it is possible for a preceding RELEASE to pass a
-later ACQUIRE from the viewpoint of the CPU, but not from the viewpoint
-of the compiler. Note that deadlocks cannot be introduced by this
-interchange because if such a deadlock threatened, the RELEASE would
-simply complete.
-
-If it is necessary for a RELEASE-ACQUIRE pair to produce a full barrier, the
-ACQUIRE can be followed by an smp_mb__after_unlock_lock() invocation. This
-will produce a full barrier if either (a) the RELEASE and the ACQUIRE are
-executed by the same CPU or task, or (b) the RELEASE and ACQUIRE act on the
-same variable. The smp_mb__after_unlock_lock() primitive is free on many
-architectures. Without smp_mb__after_unlock_lock(), the critical sections
-corresponding to the RELEASE and the ACQUIRE can cross:
+When the ACQUIRE and RELEASE are a lock acquisition and release,
+respectively, this same reordering can occur if the lock's ACQUIRE and
+RELEASE are to the same lock variable, but only from the perspective of
+another CPU not holding that lock. In short, a ACQUIRE followed by an
+RELEASE may -not- be assumed to be a full memory barrier.
+
+Similarly, the reverse case of a RELEASE followed by an ACQUIRE does not
+imply a full memory barrier. If it is necessary for a RELEASE-ACQUIRE
+pair to produce a full barrier, the ACQUIRE can be followed by an
+smp_mb__after_unlock_lock() invocation. This will produce a full barrier
+if either (a) the RELEASE and the ACQUIRE are executed by the same
+CPU or task, or (b) the RELEASE and ACQUIRE act on the same variable.
+The smp_mb__after_unlock_lock() primitive is free on many architectures.
+Without smp_mb__after_unlock_lock(), the CPU's execution of the critical
+sections corresponding to the RELEASE and the ACQUIRE can cross, so that:
*A = a;
RELEASE M
@@ -1722,7 +1749,36 @@ could occur as:
ACQUIRE N, STORE *B, STORE *A, RELEASE M
-With smp_mb__after_unlock_lock(), they cannot, so that:
+It might appear that this reordering could introduce a deadlock.
+However, this cannot happen because if such a deadlock threatened,
+the RELEASE would simply complete, thereby avoiding the deadlock.
+
+ Why does this work?
+
+ One key point is that we are only talking about the CPU doing
+ the reordering, not the compiler. If the compiler (or, for
+ that matter, the developer) switched the operations, deadlock
+ -could- occur.
+
+ But suppose the CPU reordered the operations. In this case,
+ the unlock precedes the lock in the assembly code. The CPU
+ simply elected to try executing the later lock operation first.
+ If there is a deadlock, this lock operation will simply spin (or
+ try to sleep, but more on that later). The CPU will eventually
+ execute the unlock operation (which preceded the lock operation
+ in the assembly code), which will unravel the potential deadlock,
+ allowing the lock operation to succeed.
+
+ But what if the lock is a sleeplock? In that case, the code will
+ try to enter the scheduler, where it will eventually encounter
+ a memory barrier, which will force the earlier unlock operation
+ to complete, again unraveling the deadlock. There might be
+ a sleep-unlock race, but the locking primitive needs to resolve
+ such races properly in any case.
+
+With smp_mb__after_unlock_lock(), the two critical sections cannot overlap.
+For example, with the following code, the store to *A will always be
+seen by other CPUs before the store to *B:
*A = a;
RELEASE M
@@ -1730,13 +1786,18 @@ With smp_mb__after_unlock_lock(), they cannot, so that:
smp_mb__after_unlock_lock();
*B = b;
-will always occur as either of the following:
+The operations will always occur in one of the following orders:
- STORE *A, RELEASE, ACQUIRE, STORE *B
- STORE *A, ACQUIRE, RELEASE, STORE *B
+ STORE *A, RELEASE, ACQUIRE, smp_mb__after_unlock_lock(), STORE *B
+ STORE *A, ACQUIRE, RELEASE, smp_mb__after_unlock_lock(), STORE *B
+ ACQUIRE, STORE *A, RELEASE, smp_mb__after_unlock_lock(), STORE *B
If the RELEASE and ACQUIRE were instead both operating on the same lock
-variable, only the first of these two alternatives can occur.
+variable, only the first of these alternatives can occur. In addition,
+the more strongly ordered systems may rule out some of the above orders.
+But in any case, as noted earlier, the smp_mb__after_unlock_lock()
+ensures that the store to *A will always be seen as happening before
+the store to *B.
Locks and semaphores may not provide any guarantee of ordering on UP compiled
systems, and so cannot be counted on in such a situation to actually achieve
@@ -2757,7 +2818,7 @@ in that order, but, without intervention, the sequence may have almost any
combination of elements combined or discarded, provided the program's view of
the world remains consistent. Note that ACCESS_ONCE() is -not- optional
in the above example, as there are architectures where a given CPU might
-interchange successive loads to the same location. On such architectures,
+reorder successive loads to the same location. On such architectures,
ACCESS_ONCE() does whatever is necessary to prevent this, for example, on
Itanium the volatile casts used by ACCESS_ONCE() cause GCC to emit the
special ld.acq and st.rel instructions that prevent such reordering.