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SubjectRe: [RFC v3 2/6] Improve the tracking of active utilisation
On 10/11/16 10:04, Juri Lelli wrote:
> On 02/11/16 03:35, Luca Abeni wrote:
> > On Tue, 1 Nov 2016 22:46:33 +0100
> > luca abeni <luca.abeni@unitn.it> wrote:
> > [...]
> > > > > @@ -1074,6 +1161,14 @@ select_task_rq_dl(struct task_struct *p, int cpu, int sd_flag, int flags)
> > > > > }
> > > > > rcu_read_unlock();
> > > > >
> > > > > + rq = task_rq(p);
> > > > > + raw_spin_lock(&rq->lock);
> > > > > + if (hrtimer_active(&p->dl.inactive_timer)) {
> > > > > + sub_running_bw(&p->dl, &rq->dl);
> > > > > + hrtimer_try_to_cancel(&p->dl.inactive_timer);
> > > >
> > > > Can't we subtract twice if it happens that after we grabbed rq_lock the timer
> > > > fired, so it's now waiting for that lock and it goes ahead and sub_running_bw
> > > > again after we release the lock?
> > > Uhm... I somehow convinced myself that this could not happen, but I do not
> > > remember the details, sorry :(
> > I think I remember the answer now: pi_lock is acquired before invoking select_task_rq
> > and is released after invoking enqueue_task... So, if there is a pending inactive
> > timer, its handler will be executed after the task is enqueued... It will see the task
> > as RUNNING, and will not decrease the active utilisation.
> >
>
> Oh, because we do task_rq_lock() inactive_task_timer(). So, that should
> save us from the double subtract. Would you mind adding something along
> the line of what you said above as a comment for next version?
>

Mmm, wait again.

Cannot the following happen?

- inactive_timer fires and does sub_running_bw (as the task is not
RUNNING)
- another cpu does try_to_wake_up and blocks on pi_lock
- inactive timer releases both pi and rq locks (but is still executing,
let's say it is doing put_task_struct())
- try_to_wake_up goes ahead and calls select_task_rq_dl
+ it finds inactive_timer active
+ sub_running_bw again :(

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