sched: Remove useless code in yield_to()

It's impossible to enter the else branch if we have set
skip_clock_update in task_yield_fair(), as yield_to_task_fair()
 will directly return true after invoke task_yield_fair().
Signed-off-by: NMichael Wang <wangyun@linux.vnet.ibm.com>
Acked-by: NMike Galbraith <efault@gmx.de>
Signed-off-by: NPeter Zijlstra <a.p.zijlstra@chello.nl>
Link: http://lkml.kernel.org/r/4FF2925A.9060005@linux.vnet.ibm.comSigned-off-by: NIngo Molnar <mingo@kernel.org>

kernel/sched/core.c

@@ -4348,13 +4348,6 @@ bool __sched yield_to(struct task_struct *p, bool preempt)
 		 */
 		if (preempt && rq != p_rq)
 			resched_task(p_rq->curr);
-	} else {
-		/*
-		 * We might have set it in task_yield_fair(), but are
-		 * not going to schedule(), so don't want to skip
-		 * the next update.
-		 */
-		rq->skip_clock_update = 0;
 	}
 
 out: