On 29/10/20 17:38, Peter Zijlstra wrote: > 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. >
Gotcha. Thanks, and apologies for the noise. >> > + * 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;
