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Documentation/memory-barriers.txt: Need barriers() for some control dependencies

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Current compilers can "speculate" stores in the case where both legs
of the "if" statement start with identical stores.  Because the stores
are identical, the compiler knows that the store will unconditionally
execute regardless of the "if" condition, and so the compiler is within
its rights to hoist the store to precede the condition.  Such hoisting
destroys the control-dependency ordering.  This ordering can be restored
by placing a barrier() at the beginning of each leg of the "if" statement.
This commit adds this requirement to the control-dependencies section.

Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com>
This commit is contained in:
Paul E. McKenney 2014-02-12 20:19:47 -08:00
parent bbf393b0d5
commit 9b2b3bf531
1 changed files with 19 additions and 7 deletions
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26
Documentation/memory-barriers.txt
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@ -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();
}
@ -723,6 +731,10 @@ 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, and this
conditional must involve the prior load. If the compiler