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>

1 files changed, 0 insertions, 7 deletions

diff --git a/kernel/sched/core.c b/kernel/sched/core.c
index ec0f2b81b81c..c46a011ce5db 100644
--- a/kernel/sched/core.c
+++ b/kernel/sched/core.c
@@ -4348,13 +4348,6 @@ again:
                  */
                 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: