On Tue, 04 Jul 2006 09:49:14 +0200
Jes Sorensen <[email protected]> wrote:
> Keith Owens wrote:
> >> static void invalidate_bh_lrus(void)
> >> {
> >> - on_each_cpu(invalidate_bh_lru, NULL, 1, 1);
> >> + /*
> >> + * Need to hand down a copy of the mask or we wouldn't be run
> >> + * anywhere due to the original mask being cleared
> >> + */
> >> + cpumask_t mask = lru_in_use;
> >> + cpus_clear(lru_in_use);
> >> + schedule_on_each_cpu_mask(invalidate_bh_lru, NULL, mask);
> >> }
> >
> > Racy? Start with an empty lru_in_use.
> >
> > Cpu A Cpu B
> > invalidate_bh_lrus()
> > mask = lru_in_use;
> > preempted
> > block I/O
> > bh_lru_install()
> > cpu_set(cpu, lru_in_use);
> > resume
> > cpus_clear(lru_in_use);
> > schedule_on_each_cpu_mask() - does not send IPI to cpu B
>
> I guess the real question is whether the device is still valid for new
> IO by the time we hit the invalidate function. If not, and that was my,
> possibly flawed, assumption, then it's not an issue. Whatever bh's are
> left in the lrus from other devices will be handled on the next hit.
>
The problem is that we can actually lose bits in invalidate_bh_lrus().
CPU0 CPU1
mask = lru_in_use;
cpu_set(lru_in_use, 1);
cpus_clear(lru_in_use);
Now we have a stray bh which nobody knows about.
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