Use the list_is_singular(&rt_se->run_list) api instead of
'rt_se->run_list.prev != rt_se->run_list.next'.
Fix a comment by the way, and make the comment more clearly.

Signed-off-by: Hui Su <sh_...@163.com>
---
 kernel/sched/rt.c | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/kernel/sched/rt.c b/kernel/sched/rt.c
index 49ec096a8aa1..1479d00656b4 100644
--- a/kernel/sched/rt.c
+++ b/kernel/sched/rt.c
@@ -2381,7 +2381,7 @@ static inline void watchdog(struct rq *rq, struct 
task_struct *p) { }
  *
  * NOTE: This function can be called remotely by the tick offload that
  * goes along full dynticks. Therefore no local assumption can be made
- * and everything must be accessed through the @rq and @curr passed in
+ * and everything must be accessed through the @rq and @p passed in
  * parameters.
  */
 static void task_tick_rt(struct rq *rq, struct task_struct *p, int queued)
@@ -2406,11 +2406,11 @@ static void task_tick_rt(struct rq *rq, struct 
task_struct *p, int queued)
        p->rt.time_slice = sched_rr_timeslice;
 
        /*
-        * Requeue to the end of queue if we (and all of our ancestors) are not
-        * the only element on the queue
+        * Requeue to the end of rt_prio_array queue if we (and all of our
+        * ancestors) are not the only element on the rt_prio_array queue.
         */
        for_each_sched_rt_entity(rt_se) {
-               if (rt_se->run_list.prev != rt_se->run_list.next) {
+               if (!list_is_singular(&rt_se->run_list)) {
                        requeue_task_rt(rq, p, 0);
                        resched_curr(rq);
                        return;
-- 
2.25.1


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