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 -- To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/