Messages in this thread | | | From | Namhyung Kim <> | Date | Mon, 9 Jul 2012 16:07:23 +0900 | Subject | Re: [Question] sched/rt_mutex: re-enqueue_task on rt_mutex_setprio() |
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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
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