memory-barriers: Fix control-ordering no-transitivity example

The control-ordering example demonstrating lack of transitivity had
multiple problems.  This commit fixes them.

Reported-by: Nikolay Samofatov <nikolay.samofatov@gmail.com>
Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com>
Reviewed-by: Pranith Kumar <bobby.prani@gmail.com>

diff --git a/Documentation/memory-barriers.txt b/Documentation/memory-barriers.txt
index a4de88fb55f087451cf97955a6ab419b74935c4e..d67c508eb660b1f83439180afa0885504abbea2a 100644 (file)
--- a/Documentation/memory-barriers.txt
+++ b/Documentation/memory-barriers.txt
@@ -697,30 +697,36 @@ should do something like the following:
        }
 
 Finally, control dependencies do -not- provide transitivity.  This is
-demonstrated by two related examples:
+demonstrated by two related examples, with the initial values of
+x and y both being zero:
 
        CPU 0                     CPU 1
        =====================     =====================
        r1 = ACCESS_ONCE(x);      r2 = ACCESS_ONCE(y);
-       if (r1 >= 0)              if (r2 >= 0)
+       if (r1 > 0)               if (r2 > 0)
          ACCESS_ONCE(y) = 1;       ACCESS_ONCE(x) = 1;
 
        assert(!(r1 == 1 && r2 == 1));
 
 The above two-CPU example will never trigger the assert().  However,
 if control dependencies guaranteed transitivity (which they do not),
-then adding the following two CPUs would guarantee a related assertion:
+then adding the following CPU would guarantee a related assertion:
 
-       CPU 2                     CPU 3
-       =====================     =====================
-       ACCESS_ONCE(x) = 2;       ACCESS_ONCE(y) = 2;
+       CPU 2
+       =====================
+       ACCESS_ONCE(x) = 2;
+
+       assert(!(r1 == 2 && r2 == 1 && x == 2)); /* FAILS!!! */
 
-       assert(!(r1 == 2 && r2 == 2 && x == 1 && y == 1)); /* FAILS!!! */
+But because control dependencies do -not- provide transitivity, the above
+assertion can fail after the combined three-CPU example completes.  If you
+need the three-CPU example to provide ordering, you will need smp_mb()
+between the loads and stores in the CPU 0 and CPU 1 code fragments,
+that is, just before or just after the "if" statements.
 
-But because control dependencies do -not- provide transitivity, the
-above assertion can fail after the combined four-CPU example completes.
-If you need the four-CPU example to provide ordering, you will need
-smp_mb() between the loads and stores in the CPU 0 and CPU 1 code fragments.
+These two examples are the LB and WWC litmus tests from this paper:
+http://www.cl.cam.ac.uk/users/pes20/ppc-supplemental/test6.pdf and this
+site: https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html.
 
 In summary: