linux-kernel - Re: [RFC v3 2/6] Improve the tracking of active utilisation

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Date:   Wed, 2 Nov 2016 03:35:52 +0100
From:   luca abeni <luca.abeni@...tn.it>
To:     Juri Lelli <juri.lelli@....com>
Cc:     linux-kernel@...r.kernel.org,
        Peter Zijlstra <peterz@...radead.org>,
        Ingo Molnar <mingo@...hat.com>,
        Claudio Scordino <claudio@...dence.eu.com>,
        Steven Rostedt <rostedt@...dmis.org>
Subject: Re: [RFC v3 2/6] Improve the tracking of active utilisation

On Tue, 1 Nov 2016 22:46:33 +0100
luca abeni <luca.abeni@...tn.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