[PATCH v2 1/2] sched/deadline: Remove unnecessary condition in push_dl_task()

[PATCH v2 1/2] sched/deadline: Remove unnecessary condition in push_dl_task()

[Byungchul Park]

Once pick_next_pushable_dl_task(rq) returns a task, it guarantees that
the task's cpu is rq->cpu, so task_cpu(next_task) is always rq->cpu if
task == next_task. Remove the redundant condition and make code simpler.

pick_next_pushable_dl_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.

By this patch, unnecessary one branch and two LOAD operations in 'if'
statement can be avoided.

Signed-off-by: Byungchul Park <byungchul.park@lge.com>
---
 kernel/sched/deadline.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/kernel/sched/deadline.c b/kernel/sched/deadline.c
index 27737f3..ad8d577 100644
--- a/kernel/sched/deadline.c
+++ b/kernel/sched/deadline.c
@@ -1483,7 +1483,7 @@ static int push_dl_task(struct rq *rq)
 		 * then possible that next_task has migrated.
 		 */
 		task = pick_next_pushable_dl_task(rq);
-		if (task_cpu(next_task) == rq->cpu && task == next_task) {
+		if (task == next_task) {
 			/*
 			 * The task is still there. We don't try
 			 * again, some other cpu will pull it when ready.
-- 
1.9.1

[PATCH v2 2/2] sched/rt: Remove unnecessary condition in push_rt_task()

[Byungchul Park]

Once pick_next_pushable_task(rq) returns a task, it guarantees that
the task's cpu is rq->cpu, so task_cpu(next_task) is always rq->cpu if
task == next_task. Remove the redundant condition and make code simpler.

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.

By this patch, unnecessary one branch and two LOAD operations in 'if'
statement can be avoided.

Signed-off-by: Byungchul Park <byungchul.park@lge.com>
---
 kernel/sched/rt.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/kernel/sched/rt.c b/kernel/sched/rt.c
index 4101f9d..5abd9a52 100644
--- a/kernel/sched/rt.c
+++ b/kernel/sched/rt.c
@@ -1820,7 +1820,7 @@ 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
-- 
1.9.1

Re: [PATCH v2 1/2] sched/deadline: Remove unnecessary condition in push_dl_task()

[Steven Rostedt]

On Thu, 16 Feb 2017 11:05:53 +0900
Byungchul Park <byungchul.park@lge.com> wrote:

> Once pick_next_pushable_dl_task(rq) returns a task, it guarantees that
> the task's cpu is rq->cpu, so task_cpu(next_task) is always rq->cpu if
> task == next_task. Remove the redundant condition and make code simpler.
> 
> pick_next_pushable_dl_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.
> 
> By this patch, unnecessary one branch and two LOAD operations in 'if'
> statement can be avoided.
> 
> Signed-off-by: Byungchul Park <byungchul.park@lge.com>

Reviewed-by: Steven Rostedt (VMware) <rostedt@goodmis.org>

-- Steve

> ---
>  kernel/sched/deadline.c | 2 +-
>  1 file changed, 1 insertion(+), 1 deletion(-)
> 
> diff --git a/kernel/sched/deadline.c b/kernel/sched/deadline.c
> index 27737f3..ad8d577 100644
> --- a/kernel/sched/deadline.c
> +++ b/kernel/sched/deadline.c
> @@ -1483,7 +1483,7 @@ static int push_dl_task(struct rq *rq)
>  		 * then possible that next_task has migrated.
>  		 */
>  		task = pick_next_pushable_dl_task(rq);
> -		if (task_cpu(next_task) == rq->cpu && task == next_task) {
> +		if (task == next_task) {
>  			/*
>  			 * The task is still there. We don't try
>  			 * again, some other cpu will pull it when ready.

Re: [PATCH v2 2/2] sched/rt: Remove unnecessary condition in push_rt_task()

[Steven Rostedt]

On Thu, 16 Feb 2017 11:05:54 +0900
Byungchul Park <byungchul.park@lge.com> wrote:

> Once pick_next_pushable_task(rq) returns a task, it guarantees that
> the task's cpu is rq->cpu, so task_cpu(next_task) is always rq->cpu if
> task == next_task. Remove the redundant condition and make code simpler.
> 
> 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.
> 
> By this patch, unnecessary one branch and two LOAD operations in 'if'
> statement can be avoided.
> 
> Signed-off-by: Byungchul Park <byungchul.park@lge.com>
> ---
>  kernel/sched/rt.c | 2 +-
>  1 file changed, 1 insertion(+), 1 deletion(-)
> 
> diff --git a/kernel/sched/rt.c b/kernel/sched/rt.c
> index 4101f9d..5abd9a52 100644
> --- a/kernel/sched/rt.c
> +++ b/kernel/sched/rt.c
> @@ -1820,7 +1820,7 @@ 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) {

Reviewed-by: Steven Rostedt (VMware) <rostedt@goodmis.org>

-- Steve

>  			/*
>  			 * The task hasn't migrated, and is still the next
>  			 * eligible task, but we failed to find a run-queue