Re: [PATCH v3 5/6] Also record sleep start for a task group

Paul Turner
2012-05-30T11:35:46Z

On Wed, May 30, 2012 at 2:48 AM, Glauber Costa <glommer@parallels.com> wrote:
> When we're dealing with a task group, instead of a task, also record
> the start of its sleep time. Since the test agains TASK_UNINTERRUPTIBLE
> does not really make sense and lack an obvious analogous, we always
> record it as sleep_start, never block_start.
>
> Signed-off-by: Glauber Costa <glommer@parallels.com>
> CC: Peter Zijlstra <a.p.zijlstra@chello.nl>
> CC: Paul Turner <pjt@google.com>
> ---
>  kernel/sched/fair.c |    3 ++-
>  1 file changed, 2 insertions(+), 1 deletion(-)
>
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index c26fe38..d932559 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -1182,7 +1182,8 @@ dequeue_entity(struct cfs_rq *cfs_rq, struct sched_entity *se, int flags)
>                                se->statistics.sleep_start = rq_of(cfs_rq)->clock;
>                        if (tsk->state & TASK_UNINTERRUPTIBLE)
>                                se->statistics.block_start = rq_of(cfs_rq)->clock;
> -               }
> +               } else
> +                       se->statistics.sleep_start = rq_of(cfs_rq)->clock;

You can't sanely account sleep on a group entity.

Suppose you have 2 sleepers on 1 cpu: you account 1s/s of idle
Suppose you have 2 sleepers now on 2 cpus: you account 2s/s of idle

Furthermore, in the latter case when one wakes up you still continue
to accrue sleep time whereas in the former you don't.

Just don't report/collect this.

>  #endif
>        }
>
> --
> 1.7.10.2
>