linux-kernel - Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex

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Date:	Mon, 9 Jul 2012 16:07:23 +0900
From:	Namhyung Kim <namhyung@...nel.org>
To:	Peter Zijlstra <peterz@...radead.org>
Cc:	Steven Rostedt <rostedt@...dmis.org>,
	Ingo Molnar <mingo@...nel.org>,
	LKML <linux-kernel@...r.kernel.org>
Subject: Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

On Mon, Jul 9, 2012 at 3:48 PM, Peter Zijlstra <peterz@...radead.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
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