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authorPaul E. McKenney <paulmck@linux.vnet.ibm.com>2014-10-12 07:55:47 -0700
committerPaul E. McKenney <paulmck@linux.vnet.ibm.com>2014-11-13 10:34:53 -0800
commit8b19d1dead8413442ba0ff0b4e19b08f69d2f1b7 (patch)
treee2dcecc0d8057974aac7b5929f4adfa374351688
parent74860feed5ed570659e0f3852dd945be5b046038 (diff)
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documentation: Additional restriction for control dependencies
Short-circuit booleans are not defences against compilers breaking your intended control dependencies. Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com> Reviewed-by: Pranith Kumar <bobby.prani@gmail.com>
-rw-r--r--Documentation/memory-barriers.txt18
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diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index 22a969c..1073e01 100644
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -694,6 +694,24 @@ Please note once again that the stores to 'b' differ. If they were
identical, as noted earlier, the compiler could pull this store outside
of the 'if' statement.
+You must also be careful not to rely too much on boolean short-circuit
+evaluation. Consider this example:
+
+ q = ACCESS_ONCE(a);
+ if (a || 1 > 0)
+ ACCESS_ONCE(b) = 1;
+
+Because the second condition is always true, the compiler can transform
+this example as following, defeating control dependency:
+
+ q = ACCESS_ONCE(a);
+ ACCESS_ONCE(b) = 1;
+
+This example underscores the need to ensure that the compiler cannot
+out-guess your code. More generally, although ACCESS_ONCE() does force
+the compiler to actually emit code for a given load, it does not force
+the compiler to use the results.
+
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:
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