sched: Remove one division operation in find_busiest_queue() (95a79b80) · Commits · 戴 / test

Remove one division operation in find_busiest_queue() by using crosswise multiplication: wl_i / power_i > wl_j / power_j := wl_i * power_j > wl_j * power_i Signed-off-by:

Joonsoo Kim <iamjoonsoo.kim@lge.com> [ Expanded the changelog. ] Signed-off-by:

Peter Zijlstra <peterz@infradead.org> Link: http://lkml.kernel.org/r/1375778203-31343-2-git-send-email-iamjoonsoo.kim@lge.com Signed-off-by:

Ingo Molnar <mingo@kernel.org>

kernel/sched/fair.c

+9 −5

Original line number	Diff line number	Diff line
		@@ -4968,7 +4968,7 @@ static struct rq find_busiest_queue(struct lb_env env,
		struct sched_group *group)
		{
		struct rq busiest = NULL, rq;
		unsigned long max_load = 0;
		unsigned long busiest_load = 0, busiest_power = 1;
		int i;

		for_each_cpu(i, sched_group_cpus(group)) {
		@@ -4998,11 +4998,15 @@ static struct rq find_busiest_queue(struct lb_env env,
		* the weighted_cpuload() scaled with the cpu power, so that
		* the load can be moved away from the cpu that is potentially
		* running at a lower capacity.
		*/
		wl = (wl * SCHED_POWER_SCALE) / power;

		if (wl > max_load) {
		max_load = wl;
		*
		* Thus we're looking for max(wl_i / power_i), crosswise
		* multiplication to rid ourselves of the division works out
		* to: wl_i * power_j > wl_j * power_i; where j is our
		* previous maximum.
		*/
		if (wl * busiest_power > busiest_load * power) {
		busiest_load = wl;
		busiest_power = power;
		busiest = rq;
		}
		}

Admin message