Re: [RFC v2 3/7] Improve the tracking of active utilisation

From: Peter Zijlstra
Date: Tue Apr 05 2016 - 15:31:13 EST


On Tue, Apr 05, 2016 at 09:24:24PM +0200, luca abeni wrote:
> On Tue, 5 Apr 2016 20:02:52 +0200
> Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
>
> > On Tue, Apr 05, 2016 at 07:56:57PM +0200, luca abeni wrote:
> >
> > > > > + migrate_active = hrtimer_active(&p->dl.inactive_timer);
> > > > > + if (migrate_active)
> > > > > + sub_running_bw(&p->dl, &rq->dl);
> > > > > + raw_spin_unlock(&rq->lock);
> > > >
> > > > At this point task_rq() is still the above rq, so if the inactive timer
> > > > hits here it will lock this rq and subtract the running bw here _again_,
> > > > right?
> > > I think it will see the task state as TASK_RUNNING, so it will do nothing.
> > > Or it will cancelled later when the task is enqueued... I'll double check this.
> >
> > Right, so this is select_task_rq_dl(), we run this in wakeups, before
> > TASK_RUNNING.
>
> Sigh... I knew I was missing something here... :(
> So, I think the solution here is to use double_lock_balance() (or something
> like that) to take both the rq locks so that the inactive timer handler cannot
> run between sub_running_bw() and add_running_bw()... I'll try this.

I'm not sure that'll fix it, because after you unlock both again, we can
hit after, and there task_rq() will still be the first rq, not the
second. So we again subtract twice from the old rq.

Only after __set_task_cpu()'s store to task_thread_info(p)->cpu will the
timer hit the new rq.

And you cannot hold a lock over that..