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: Michael Wang <wangyun@linux.vnet.ibm.com>
Acked-by: Mike Galbraith <efault@gmx.de>
Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl>
Link: http://lkml.kernel.org/r/4FF2925A.9060005@linux.vnet.ibm.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>

diff --git a/kernel/sched/core.c b/kernel/sched/core.c
index ec0f2b81b81ce13f5a7d8cb8b8a23bf075db89c0..c46a011ce5db89c76a0d466a0f7a965204a86aec 100644 (file)
--- a/kernel/sched/core.c
+++ b/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: