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

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;

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