Re: [PATCH v4 14/19] sched, lockdep: Annotate ->pi_lock recursion

[Peter Zijlstra]

On Thu, Oct 29, 2020 at 04:27:16PM +0000, Valentin Schneider wrote:
> 
> On 23/10/20 11:12, Peter Zijlstra wrote:
> > @@ -2617,6 +2618,20 @@ void sched_set_stop_task(int cpu, struct
> >               sched_setscheduler_nocheck(stop, SCHED_FIFO, &param);
> >
> >               stop->sched_class = &stop_sched_class;
> > +
> > +		/*
> > +		 * The PI code calls rt_mutex_setprio() with ->pi_lock held to
> > +		 * adjust the effective priority of a task. As a result,
> > +		 * rt_mutex_setprio() can trigger (RT) balancing operations,
> > +		 * which can then trigger wakeups of the stop thread to push
> > +		 * around the current task.
> > +		 *
> > +		 * The stop task itself will never be part of the PI-chain, it
> > +		 * never blocks, therefore that ->pi_lock recursion is safe.
> 
> Isn't it that the stopper task can only run when preemption is re-enabled,
> and the ->pi_lock is dropped before then?
> 
> If we were to have an SCA-like function that would kick the stopper but
> "forget" to release the pi_lock, then we would very much like lockdep to
> complain, right? Or is that something else entirely?

You've forgotten the other, and original, purpose of ->pi_lock, guarding
the actual PI chain. Please have a look at rt_mutex_adjust_prio_chain()
and its comment.

But no, this isn't about running, this is about doing an actual wakeup
(of the stopper task) while holding an ->pi_lock instance (guaranteed
not the stopper task's). And since wakeup will take ->pi_lock, lockdep
will get all whiny about ->pi_lock self recursion.

> > +		 * Tell lockdep about this by placing the stop->pi_lock in its
> > +		 * own class.
> > +		 */
> > +		lockdep_set_class(&stop->pi_lock, &stop_pi_lock);
> >       }
> >
> >       cpu_rq(cpu)->stop = stop;