On Tue, 1 Nov 2016 22:46:33 +0100
luca abeni <luca.ab...@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.
Does this make sense?
Luca
