sched/rt: Remove unnecessary condition in push_rt_task()

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. Remove the redundant condition and make the code simpler.

This way one unnecessary branch and two LOAD operations can be avoided.
Signed-off-by: Byungchul Park <byungchul.park@lge.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Reviewed-by: Steven Rostedt (VMware) <rostedt@goodmis.org>
Reviewed-by: Juri Lelli <juri.lelli@arm.com>
Reviewed-by: Daniel Bristot de Oliveira <bristot@redhat.com>
Cc: <kernel-team@lge.com>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Link: http://lkml.kernel.org/r/1494551143-22219-1-git-send-email-byungchul.park@lge.comSigned-off-by: Ingo Molnar <mingo@kernel.org>

kernel/sched/rt.c

@@ -1819,7 +1819,7 @@ 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