On Tue, May 29, 2018 at 3:49 PM Alan Stern <stern@xxxxxxxxxxxxxxxxxxx>
wrote:
> Putting this into herd would be extremely difficult, if not impossible,
> because it involves analyzing code that was not executed.
Does it?
Can't we simplify the whole sequence as basically
A
if (!B)
D
for that "not B" case, and just think about that. IOW, let's ignore the
whole "not executed" code.
If B depends on A like you state, then that already implies that the write
in D cannot come before the read of A.
You fundamentally cannot do a conditional write before the read that the
write condition depends on. So *any* write after a conditional is dependent
on the read.
So the existence of C - whether it has a barrier or not - is entirely
immaterial at run-time.
Now, the *compiler* can use the whole existence of that memory barrier in C
to determine whether it can re-order the write to D or not, of course, but
that's a separate issue, and then the whole "code that isn't executed" is
not the issue any more. The compiler obviously sees all code, whether
executing or not.
Or am I being stupid and missing something entirely? That's possible.
Linus
