On Wed, 31 Jul 2019 18:32:47 +0100
Dietmar Eggemann <[email protected]> wrote:
[...]
> >>>> static void dequeue_dl_entity(struct sched_dl_entity *dl_se)
> >>>> {
> >>>> + if (!on_dl_rq(dl_se))
> >>>> + return;
> >>>
> >>> Why allow double dequeue instead of WARN?
> >>
> >> As I was saying to Valentin, it can currently happen that a task
> >> could have already been dequeued by update_curr_dl()->throttle
> >> called by dequeue_task_dl() before calling __dequeue_task_dl(). Do
> >> you think we should check for this condition before calling into
> >> dequeue_dl_entity()?
> >
> > Yes, that's what ->dl_throttled is for, right? And !->dl_throttled
> > && !on_dl_rq() is a BUG.
>
> OK, I will add the following snippet to the patch.
> Although it's easy to provoke a situation in which DL tasks are
> throttled, I haven't seen a throttling happening when the task is
> being dequeued.
This is a not-so-common situation, that can happen with periodic tasks
(a-la rt-app) blocking on clock_nanosleep() (or similar) after
executing for an amount of time comparable with the SCHED_DEADLINE
runtime.
It might happen that the task consumed a little bit more than the
remaining runtime (but has not been throttled yet, because the
accounting happens at every tick)... So, when dequeue_task_dl() invokes
update_task_dl() the runtime becomes negative and the task is throttled.
This happens infrequently, but if you try rt-app tasksets with multiple
tasks and execution times near to the runtime you will see it
happening, sooner or later.
[...]
> @@ -1592,6 +1591,10 @@ static void __dequeue_task_dl(struct rq *rq,
> struct task_struct *p) static void dequeue_task_dl(struct rq *rq,
> struct task_struct *p, int flags) {
> update_curr_dl(rq);
> +
> + if (p->dl.dl_throttled)
> + return;
Sorry, I missed part of the previous discussion, so maybe I am missing
something... But I suspect this "return" might be wrong (you risk to
miss a call to task_non_contending(), coming later in this function).
Maybe you cound use
if (!p->dl_throttled)
__dequeue_task_dl(rq, p)
Or did I misunderstand something?
Thanks,
Luca
