Hi,
On Mon, Apr 6, 2020 at 7:14 PM Ming Lei <ming.lei@xxxxxxxxxx> wrote:
>
> On Sun, Apr 05, 2020 at 09:26:39AM -0700, Doug Anderson wrote:
> > Hi,
> >
> > On Sun, Apr 5, 2020 at 2:15 AM Ming Lei <ming.lei@xxxxxxxxxx> wrote:
> > >
> > > @@ -103,6 +104,9 @@ static void blk_mq_do_dispatch_sched(struct blk_mq_hw_ctx *hctx)
> > > rq = e->type->ops.dispatch_request(hctx);
> > > if (!rq) {
> > > blk_mq_put_dispatch_budget(hctx);
> > > +
> > > + if (e->type->ops.has_work && e->type->ops.has_work(hctx))
> > > + blk_mq_delay_run_hw_queue(hctx, BLK_MQ_BUDGET_DELAY);
> >
> > To really close the race, don't we need to run all the queues
> > associated with the hctx? I haven't traced it through, but I've been
> > assuming that the multiple "hctx"s associated with the same queue will
> > have the same budget associated with them and thus they can block each
> > other out.
>
> Yeah, we should run all hctxs which share the same budget space.
>
> Also, in theory, we don't have to add the delay, however BFQ may plug the
> dispatch for 9 ms, so looks delay run queue is required.
I have posted up v3. A few notes:
* Since we should run all "hctxs" I took out the check for has_work()
before kicking the queue.
* As far as I can tell the theoretical race happens for _anyone_ who
puts budget, so I added it to blk_mq_put_dispatch_budget(). Feel free
to shout at me in response to v3 if you believe I'm wrong about that.
Thanks for all your reviews and suggestions so far!
-Doug
