Re: [PATCH-v3 1/4] idr: Percpu ida

[Kent Overstreet]

On Wed, Aug 28, 2013 at 01:23:32PM -0700, Andrew Morton wrote:
> 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.

It's the same semantics as a mempool, really - it'll succeed when a tag
gets freed.  If we are sleeping then there isn't really anything else we
can do, there isn't anything we're trying in the __GFP_WAIT case that
we're not trying in the GFP_NOWAIT case.
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