Commit ebff09a6 authored by Paul E. McKenney's avatar Paul E. McKenney Committed by Ingo Molnar

locking/Documentation: Clarify limited control-dependency scope

Nothing in the control-dependencies section of memory-barriers.txt
says that control dependencies don't extend beyond the end of the
if-statement containing the control dependency.  Worse yet, in many
situations, they do extend beyond that if-statement.  In particular,
the compiler cannot destroy the control dependency given proper use of
READ_ONCE() and WRITE_ONCE().  However, a weakly ordered system having
a conditional-move instruction provides the control-dependency guarantee
only to code within the scope of the if-statement itself.

This commit therefore adds words and an example demonstrating this
limitation of control dependencies.
Reported-by: default avatarWill Deacon <will.deacon@arm.com>
Signed-off-by: default avatarPaul E. McKenney <paulmck@linux.vnet.ibm.com>
Acked-by: default avatarPeter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Cc: corbet@lwn.net
Cc: linux-arch@vger.kernel.org
Cc: linux-doc@vger.kernel.org
Link: http://lkml.kernel.org/r/20160615230817.GA18039@linux.vnet.ibm.comSigned-off-by: default avatarIngo Molnar <mingo@kernel.org>
parent b316ff78
......@@ -806,6 +806,41 @@ out-guess your code. More generally, although READ_ONCE() does force
the compiler to actually emit code for a given load, it does not force
the compiler to use the results.
In addition, control dependencies apply only to the then-clause and
else-clause of the if-statement in question. In particular, it does
not necessarily apply to code following the if-statement:
q = READ_ONCE(a);
if (q) {
WRITE_ONCE(b, p);
} else {
WRITE_ONCE(b, r);
}
WRITE_ONCE(c, 1); /* BUG: No ordering against the read from "a". */
It is tempting to argue that there in fact is ordering because the
compiler cannot reorder volatile accesses and also cannot reorder
the writes to "b" with the condition. Unfortunately for this line
of reasoning, the compiler might compile the two writes to "b" as
conditional-move instructions, as in this fanciful pseudo-assembly
language:
ld r1,a
ld r2,p
ld r3,r
cmp r1,$0
cmov,ne r4,r2
cmov,eq r4,r3
st r4,b
st $1,c
A weakly ordered CPU would have no dependency of any sort between the load
from "a" and the store to "c". The control dependencies would extend
only to the pair of cmov instructions and the store depending on them.
In short, control dependencies apply only to the stores in the then-clause
and else-clause of the if-statement in question (including functions
invoked by those two clauses), not to code following that if-statement.
Finally, control dependencies do -not- provide transitivity. This is
demonstrated by two related examples, with the initial values of
x and y both being zero:
......@@ -869,6 +904,12 @@ In summary:
atomic{,64}_read() can help to preserve your control dependency.
Please see the COMPILER BARRIER section for more information.
(*) Control dependencies apply only to the then-clause and else-clause
of the if-statement containing the control dependency, including
any functions that these two clauses call. Control dependencies
do -not- apply to code following the if-statement containing the
control dependency.
(*) Control dependencies pair normally with other types of barriers.
(*) Control dependencies do -not- provide transitivity. If you
......
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment