[Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

[Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

[Namhyung Kim]

Hi,

I have a question on the code below:

void rt_mutex_setprio(struct task_struct *p, int prio)
{
        ...
	if (on_rq)
		enqueue_task(rq, p, oldprio < prio ? ENQUEUE_HEAD : 0);

When enqueueing @p with new @prio, it seems put @p at the head of a
rq if appropriate. I guess it's the case of boosting @p with higher
priority, right? So Should the conditional be a reverse form (provided
that less number means higher priority)? Please shed some light on me.

Thanks,
Namhyung

Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

[Steven Rostedt]

On Sat, 2012-07-07 at 14:44 +0900, Namhyung Kim wrote:
> Hi,
> 
> I have a question on the code below:
> 
> void rt_mutex_setprio(struct task_struct *p, int prio)
> {
>         ...
> 	if (on_rq)
> 		enqueue_task(rq, p, oldprio < prio ? ENQUEUE_HEAD : 0);
> 
> When enqueueing @p with new @prio, it seems put @p at the head of a
> rq if appropriate. I guess it's the case of boosting @p with higher
> priority, right?

Actually, no. We put @p at the head of the queue when unboosting. If a
task is going from a high priority into a lower priority, it is still
treated as "important" for that priority, and is put to the front of the
queue (it was just higher than everything else on that queue). But if we
are boosting a task from a low priority, why put it to the head of other
tasks of its new priority, when those tasks were just higher than this
task, and this task is now just an "equal".

-- Steve

>  So Should the conditional be a reverse form (provided
> that less number means higher priority)? Please shed some light on me.
> 
> Thanks,
> Namhyung

Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

[Namhyung Kim]

On Sat, 07 Jul 2012 21:29:19 -0400, Steven Rostedt wrote:
> On Sat, 2012-07-07 at 14:44 +0900, Namhyung Kim wrote:
>> Hi,
>> 
>> I have a question on the code below:
>> 
>> void rt_mutex_setprio(struct task_struct *p, int prio)
>> {
>>         ...
>> 	if (on_rq)
>> 		enqueue_task(rq, p, oldprio < prio ? ENQUEUE_HEAD : 0);
>> 
>> When enqueueing @p with new @prio, it seems put @p at the head of a
>> rq if appropriate. I guess it's the case of boosting @p with higher
>> priority, right?
>
> Actually, no. We put @p at the head of the queue when unboosting. If a
> task is going from a high priority into a lower priority, it is still
> treated as "important" for that priority, and is put to the front of the
> queue (it was just higher than everything else on that queue). But if we
> are boosting a task from a low priority, why put it to the head of other
> tasks of its new priority, when those tasks were just higher than this
> task, and this task is now just an "equal".

Thanks for the explanation. (Isn't it worth getting commented?) :)

Thanks,
Namhyung

Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

[Peter Zijlstra]

On Mon, 2012-07-09 at 09:50 +0900, Namhyung Kim wrote:
> On Sat, 07 Jul 2012 21:29:19 -0400, Steven Rostedt wrote:
> > On Sat, 2012-07-07 at 14:44 +0900, Namhyung Kim wrote:
> >> Hi,
> >> 
> >> I have a question on the code below:
> >> 
> >> void rt_mutex_setprio(struct task_struct *p, int prio)
> >> {
> >>         ...
> >> 	if (on_rq)
> >> 		enqueue_task(rq, p, oldprio < prio ? ENQUEUE_HEAD : 0);
> >> 
> >> When enqueueing @p with new @prio, it seems put @p at the head of a
> >> rq if appropriate. I guess it's the case of boosting @p with higher
> >> priority, right?
> >
> > Actually, no. We put @p at the head of the queue when unboosting. If a
> > task is going from a high priority into a lower priority, it is still
> > treated as "important" for that priority, and is put to the front of the
> > queue (it was just higher than everything else on that queue). But if we
> > are boosting a task from a low priority, why put it to the head of other
> > tasks of its new priority, when those tasks were just higher than this
> > task, and this task is now just an "equal".
> 
> Thanks for the explanation. (Isn't it worth getting commented?) :)

Possibly, note that this part is well spec'ed by POSIX, see 

http://pubs.opengroup.org/onlinepubs/009695299/functions/xsh_chap02_08.html

SCHED_FIFO.8

Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

[Namhyung Kim]

On Mon, Jul 9, 2012 at 3:48 PM, Peter Zijlstra <peterz@infradead.org> wrote:
> On Mon, 2012-07-09 at 09:50 +0900, Namhyung Kim wrote:
>> On Sat, 07 Jul 2012 21:29:19 -0400, Steven Rostedt wrote:
>> > On Sat, 2012-07-07 at 14:44 +0900, Namhyung Kim wrote:
>> >> Hi,
>> >>
>> >> I have a question on the code below:
>> >>
>> >> void rt_mutex_setprio(struct task_struct *p, int prio)
>> >> {
>> >>         ...
>> >>    if (on_rq)
>> >>            enqueue_task(rq, p, oldprio < prio ? ENQUEUE_HEAD : 0);
>> >>
>> >> When enqueueing @p with new @prio, it seems put @p at the head of a
>> >> rq if appropriate. I guess it's the case of boosting @p with higher
>> >> priority, right?
>> >
>> > Actually, no. We put @p at the head of the queue when unboosting. If a
>> > task is going from a high priority into a lower priority, it is still
>> > treated as "important" for that priority, and is put to the front of the
>> > queue (it was just higher than everything else on that queue). But if we
>> > are boosting a task from a low priority, why put it to the head of other
>> > tasks of its new priority, when those tasks were just higher than this
>> > task, and this task is now just an "equal".
>>
>> Thanks for the explanation. (Isn't it worth getting commented?) :)
>
> Possibly, note that this part is well spec'ed by POSIX, see
>
> http://pubs.opengroup.org/onlinepubs/009695299/functions/xsh_chap02_08.html
>
> SCHED_FIFO.8

Thanks for the pointer. I need to educate myself a lot more!

Thanks,
Namhyung