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

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Message-ID: <87obnp92r8.fsf@sejong.aot.lge.com>
Date:	Mon, 09 Jul 2012 09:50:19 +0900
From:	Namhyung Kim <namhyung@...nel.org>
To:	Steven Rostedt <rostedt@...dmis.org>
Cc:	Ingo Molnar <mingo@...nel.org>,
	Peter Zijlstra <peterz@...radead.org>,
	LKML <linux-kernel@...r.kernel.org>
Subject: Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio()

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

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