Re: [PATCH 5/8] sched,cfs: use explicit cfs_rq of parent se helper

[Dietmar Eggemann]

On 6/12/19 9:32 PM, Rik van Riel wrote:
> Use an explicit "cfs_rq of parent sched_entity" helper in a few
> strategic places, where cfs_rq_of(se) may no longer point at the
> right runqueue once we flatten the hierarchical cgroup runqueues.
> 
> No functional change.
> 
> Signed-off-by: Rik van Riel <riel@xxxxxxxxxxx>
> ---
>  kernel/sched/fair.c | 17 +++++++++++++----
>  1 file changed, 13 insertions(+), 4 deletions(-)
> 
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index dcc521d251e3..c6ede2ecc935 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -275,6 +275,15 @@ static inline struct cfs_rq *group_cfs_rq(struct sched_entity *grp)
>  	return grp->my_q;
>  }
>  
> +/* runqueue owned by the parent entity */
> +static inline struct cfs_rq *group_cfs_rq_of_parent(struct sched_entity *se)
> +{
> +	if (se->parent)
> +		return group_cfs_rq(se->parent);
> +
> +	return &cfs_rq_of(se)->rq->cfs;

The function name and the description is not 100% correct. For tasks
running naturally (not in a flattened taskgroup) in the root taskgroup
or for the se representing a first level taskgroup (e.g. /tg1 (with
se->depth = 0)) it returns the root cfs_rq or easier se->cfs_rq.

So you could replace

    return &cfs_rq_of(se)->rq->cfs;

with

    return se->cfs_rq;

or

    return cfs_rq_of(se);

I guess a crucial point to understand is that you do need both,
cfs_rq_of(se) to access the flattened and group_cfs_rq_of_parent(se) to
access the hierarchical world.

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