Messages in this thread Patch in this message | | | Date | Sat, 21 May 2011 23:28:19 +0800 | Subject | [PATCH v0] sched: change how run-queue is selected for RT task | From | Hillf Danton <> |
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When selecting run-queue for a given RT task, we have to take a few factors, such as task priority and CPU cache affinity, into consideration. In this work, a simpler method is proposed, which is focusing on the relation between the current run-queue of the given task and the given run-queue.
If the current run-queue of task is the given run-queue, the run-queue of task keeps unchanged, so the CPU cache affinities of both task and the current task of run-queue remain unchanged. Then there are at least two tasks competing one CPU, and in the worst case that both competitors are RT tasks the victim will be selected and processed by pusher later.
On other hand, if the current run-queue of task is different from the given run-queue, task is simply delivered to its current run-queue, since pusher is always willing to do hard works.
In summary, the burden of RT task is always processed first by the pusher of its current run-queue.
Signed-off-by: Hillf Danton <dhillf@gmail.com> ---
--- a/kernel/sched_rt.c 2011-04-27 11:48:50.000000000 +0800 +++ b/kernel/sched_rt.c 2011-05-21 22:19:52.000000000 +0800 @@ -998,14 +998,12 @@ select_task_rq_rt(struct rq *rq, struct * * For equal prio tasks, we just let the scheduler sort it out. */ - if (unlikely(rt_task(rq->curr)) && - (rq->curr->rt.nr_cpus_allowed < 2 || - rq->curr->prio < p->prio) && - (p->rt.nr_cpus_allowed > 1)) { - int cpu = find_lowest_rq(p);
- return (cpu == -1) ? task_cpu(p) : cpu; - } + if (task_cpu(p) == rq->cpu) + return rq->cpu; + + if (likely(!rt_task(rq->curr))) + return rq->cpu;
/* * Otherwise, just let it ride on the affined RQ and the
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