On Wed, 2012-10-31 at 20:04 +0900, anish kumar wrote:
> > This now does:
> >
> > CPU 1 CPU 2
> > ----- -----
> > (flags = 0)
> > cmpxchg(flags, 0, IRQ_WORK_FLAGS)
> > (flags = 3)
> > [...]
We can still add here:
(fetch flags)
> > xchg(&flags, IRQ_WORK_BUSY)
> > (flags = 2)
> > func()
> > oflags = cmpxchg(&flags, flags,
> > nflags);
Then the cmpxchg() would fail, and oflags would be 2
> > (sees flags = 2)
> > if (flags & IRQ_WORK_PENDING)
This should be:
if (oflags & IRQ_WORK_PENDING)
> > (not true)
> > (loop)
> >
> > cmpxchg(flags, 2, 0);
> > (flags = 2)
This should be:
(flags = 0)
as we described the previous cmpxchg() as failing, flags would still be
2 before this cmpxchg(), and this one would succeed.
-- Steve
> > flags = 3
> >
> >
> >
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