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SubjectRe: [RFC PATCH] sched/core: Fix premature p->migration_pending completion
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On 04/02/21 15:30, Qais Yousef wrote:
> On 02/03/21 18:59, Valentin Schneider wrote:
>> On 03/02/21 17:23, Qais Yousef wrote:
>> > On 01/27/21 19:30, Valentin Schneider wrote:
>> >> Initial conditions:
>> >> victim.cpus_mask = {CPU0, CPU1}
>> >>
>> >> CPU0 CPU1 CPU<don't care>
>> >>
>> >> switch_to(victim)
>> >> set_cpus_allowed(victim, {CPU1})
>> >> kick CPU0 migration_cpu_stop({.dest_cpu = CPU1})
>> >> switch_to(stopper/0)
>> >> // e.g. CFS load balance
>> >> move_queued_task(CPU0, victim, CPU1);
>> >> switch_to(victim)
>> >> set_cpus_allowed(victim, {CPU0});
>> >> task_rq_unlock();
>> >> migration_cpu_stop(dest_cpu=CPU1)
>> >
>> > This migration stop is due to set_cpus_allowed(victim, {CPU1}), right?
>> >
>>
>> Right
>>
>> >> task_rq(p) != rq && pending
>> >> kick CPU1 migration_cpu_stop({.dest_cpu = CPU1})
>> >>
>> >> switch_to(stopper/1)
>> >> migration_cpu_stop(dest_cpu=CPU1)
>> >
>> > And this migration stop is due to set_cpus_allowed(victim, {CPU0}), right?
>> >
>>
>> Nein! This is a retriggering of the "current" stopper (triggered by
>> set_cpus_allowed(victim, {CPU1})), see the tail of that
>>
>> else if (dest_cpu < 0 || pending)
>>
>> branch in migration_cpu_stop(), is what I'm trying to hint at with that
>>
>> task_rq(p) != rq && pending
>
> Okay I see. But AFAIU, the work will be queued in order. So we should first
> handle the set_cpus_allowed_ptr(victim, {CPU0}) before the retrigger, no?
>
> So I see migration_cpu_stop() running 3 times
>
> 1. because of set_cpus_allowed(victim, {CPU1}) on CPU0
> 2. because of set_cpus_allowed(victim, {CPU0}) on CPU1
> 3. because of retrigger of '1' on CPU0
>

On that 'CPU<don't care>' lane, I intentionally included task_rq_unlock()
but not 'kick CPU1 migration_cpu_stop({.dest_cpu = CPU0})'. IOW, there is
nothing in that trace that queues a stopper work for 2. - it *will* happen
at some point, but harm will already have been done.

The migrate_task_to() example is potentially worse, because it doesn't rely
on which stopper work gets enqueued first - only that an extra affinity
change happens before the first stopper work grabs the pi_lock and completes.

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