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SubjectRe: [PATCH 5/8] sched,cfs: use explicit cfs_rq of parent se helper
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On 6/20/19 6:29 PM, Rik van Riel wrote:
> On Thu, 2019-06-20 at 18:23 +0200, Dietmar Eggemann wrote:
>> On 6/12/19 9:32 PM, Rik van Riel wrote:

[...]

>>> @@ -7779,7 +7788,7 @@ static void update_cfs_rq_h_load(struct
>>> cfs_rq *cfs_rq)
>>>
>>> WRITE_ONCE(cfs_rq->h_load_next, NULL);
>>> for_each_sched_entity(se) {
>>> - cfs_rq = cfs_rq_of(se);
>>> + cfs_rq = group_cfs_rq_of_parent(se);
>>
>> Why do you change this here? task_se_h_load() calls
>> update_cfs_rq_h_load() with cfs_rq = group_cfs_rq_of_parent(se)
>> because
>> the task might not be on the cfs_rq yet.
>
> Because patch 6 points cfs_rq_of(se) at the CPU's top level
> cfs_rq for every task se ...
>
> ... but I guess since I have not changed where the cfs_rq_of
> points for cgroup sched_entities, this change is not necessary
> at this time, and I should be able to go without it, in this
> function.

IMHO, since you only change set_task_rq() (p->se.cfs_rq =
&cpu_rq(cpu)->cfs instead of tg->cfs_rq[cpu] in 8/8) (used for a task)
and not init_tg_cfs_entry() (used for a taskgroup), 'cfs_rq_of(se) ==
se->parent->my_q' should still be true in update_cfs_rq_h_load().

update_cfs_rq_h_load() only deals with se's representing taskgroups. So
cfs_rq_of(se) and group_cfs_rq_of_parent(se) should deliver the same
result for these se's.

>> But inside update_cfs_rq_h_load() the first se is derived from
>> cfs_rq->tg->se[cpu_of(rq)] so in the first for_each_sched_entity()
>> loop
>> we should still start with group_cfs_rq() (se->my_q) ?

Here I was wrong. The first loop did use cfs_rq_of() and not group_cfs_rq().
[...]

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