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, ¶m); > > > > 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;
