On Wed, 28 Aug 2013 12:53:17 -0700 Kent Overstreet <kmo@xxxxxxxxxxxxx> wrote:
> > > + while (1) {
> > > + spin_lock(&pool->lock);
> > > +
> > > + /*
> > > + * prepare_to_wait() must come before steal_tags(), in case
> > > + * percpu_ida_free() on another cpu flips a bit in
> > > + * cpus_have_tags
> > > + *
> > > + * global lock held and irqs disabled, don't need percpu lock
> > > + */
> > > + prepare_to_wait(&pool->wait, &wait, TASK_UNINTERRUPTIBLE);
> > > +
> > > + if (!tags->nr_free)
> > > + alloc_global_tags(pool, tags);
> > > + if (!tags->nr_free)
> > > + steal_tags(pool, tags);
> > > +
> > > + if (tags->nr_free) {
> > > + tag = tags->freelist[--tags->nr_free];
> > > + if (tags->nr_free)
> > > + set_bit(smp_processor_id(),
> > > + pool->cpus_have_tags);
> > > + }
> > > +
> > > + spin_unlock(&pool->lock);
> > > + local_irq_restore(flags);
> > > +
> > > + if (tag >= 0 || !(gfp & __GFP_WAIT))
> > > + break;
> > > +
> > > + schedule();
> > > +
> > > + local_irq_save(flags);
> > > + tags = this_cpu_ptr(pool->tag_cpu);
> > > + }
> >
> > What guarantees that this wait will terminate?
>
> It seems fairly clear to me from the break statement a couple lines up;
> if we were passed __GFP_WAIT we terminate iff we succesfully allocated a
> tag. If we weren't passed __GFP_WAIT we never actually sleep.
OK ;) Let me rephrase. What guarantees that a tag will become available?
If what we have here is an open-coded __GFP_NOFAIL then that is
potentially problematic.
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