On Sun, Feb 21, 2016 at 4:57 AM, Paul E. McKenney <paul...@linux.vnet.ibm.com> wrote: > On Sat, Feb 20, 2016 at 03:01:08PM +0900, SeongJae Park wrote: >> There is wrong comment in example for compiler store omit behavior. It >> shows example of the problem and than problem solved version code. >> However, the comment in the solved version is still same with not solved >> version. Fix the wrong statement with this commit. >> >> Signed-off-by: SeongJae Park <sj38.p...@gmail.com> > > Hmmm... The code between the two stores of zero to "a" is intended to > remain the same in the broken and fixed versions. So the only change > is from "a = 0" to "WRITE_ONCE(a, 0)". Note that it is some other > CPU that did the third store to "a".
Agree, of course. > > Or am I missing your point here? My point is about the comment. I thought the comment in broken version is saying "Below line(a = 0) says it will store to variable 'a', but it will not in actual because a compiler can omit it". However, in fixed version, because the compiler cannot omit the store now, I thought the comment also should be changed to say the difference between broken and fixed version. If I am understanding anything wrong, please let me know. Thanks, SeongJae Park > > Thanx, Paul > >> --- >> Documentation/memory-barriers.txt | 2 +- >> 1 file changed, 1 insertion(+), 1 deletion(-) >> >> diff --git a/Documentation/memory-barriers.txt >> b/Documentation/memory-barriers.txt >> index 061ff29..b4754c7 100644 >> --- a/Documentation/memory-barriers.txt >> +++ b/Documentation/memory-barriers.txt >> @@ -1471,7 +1471,7 @@ of optimizations: >> wrong guess: >> >> WRITE_ONCE(a, 0); >> - /* Code that does not store to variable a. */ >> + /* Code that does store to variable a. */ >> WRITE_ONCE(a, 0); >> >> (*) The compiler is within its rights to reorder memory accesses unless >> -- >> 1.9.1 >> >
