[PATCH v2 2/2] sched/rt: Remove unnecessary condition in push_rt_task()

[Byungchul Park <byungchul.park@lge.com>]

Once pick_next_pushable_task(rq) returns a task, it guarantees that
the task's cpu is rq->cpu, so task_cpu(next_task) is always rq->cpu if
task == next_task. Remove the redundant condition and make code simpler.

pick_next_pushable_task(rq) has BUG_ON(rq_cpu != task_cpu(task)) when
it returns a task other than NULL, which means that task_cpu(task) must
be rq->cpu. So if task == next_task, then task_cpu(next_task) must be
rq->cpu as well.

By this patch, unnecessary one branch and two LOAD operations in 'if'
statement can be avoided.

Signed-off-by: Byungchul Park <byungchul.park@lge.com>
---
 kernel/sched/rt.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/kernel/sched/rt.c b/kernel/sched/rt.c
index 4101f9d..5abd9a52 100644
--- a/kernel/sched/rt.c
+++ b/kernel/sched/rt.c
@@ -1820,7 +1820,7 @@ static int push_rt_task(struct rq *rq)
 		 * pushing.
 		 */
 		task = pick_next_pushable_task(rq);
-		if (task_cpu(next_task) == rq->cpu && task == next_task) {
+		if (task == next_task) {
 			/*
 			 * The task hasn't migrated, and is still the next
 			 * eligible task, but we failed to find a run-queue
-- 
1.9.1