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

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Date:	Mon, 09 Jul 2012 08:48:35 +0200
From:	Peter Zijlstra <peterz@...radead.org>
To:	Namhyung Kim <namhyung@...nel.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, 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
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