提交 49ee5768 编写于 作者: P Peter Zijlstra 提交者: Ingo Molnar

sched/core: Optimize pick_next_task() for idle_sched_class

Steve noticed that when we switch from IDLE to SCHED_OTHER we fail to
take the shortcut, even though all runnable tasks are of the fair
class, because prev->sched_class != &fair_sched_class.

Since I reworked the put_prev_task() stuff, we don't really care about
prev->class here, so removing that condition will allow this case.

This increases the likely case from 78% to 98% correct for Steve's
workload.
Reported-by: NSteven Rostedt (VMware) <rostedt@goodmis.org>
Tested-by: NSteven Rostedt (VMware) <rostedt@goodmis.org>
Signed-off-by: NPeter Zijlstra (Intel) <peterz@infradead.org>
Cc: Andrew Morton <akpm@linux-foundation.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Link: http://lkml.kernel.org/r/20170119174408.GN6485@twins.programming.kicks-ass.netSigned-off-by: NIngo Molnar <mingo@kernel.org>
上级 3a09b8d4
......@@ -3321,15 +3321,14 @@ static inline void schedule_debug(struct task_struct *prev)
static inline struct task_struct *
pick_next_task(struct rq *rq, struct task_struct *prev, struct rq_flags *rf)
{
const struct sched_class *class = &fair_sched_class;
const struct sched_class *class;
struct task_struct *p;
/*
* Optimization: we know that if all tasks are in
* the fair class we can call that function directly:
*/
if (likely(prev->sched_class == class &&
rq->nr_running == rq->cfs.h_nr_running)) {
if (likely(rq->nr_running == rq->cfs.h_nr_running)) {
p = fair_sched_class.pick_next_task(rq, prev, rf);
if (unlikely(p == RETRY_TASK))
goto again;
......
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