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: Will Deacon <will.deacon@arm.com>
Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com>
Acked-by: Peter 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.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>

diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index 147ae8ec836f85666110634ff5565f4016de1d80..a4d0a99de04da76d62bc0ceebae88fd5283473e4 100644 (file)
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -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