Re: [PATCH v2] sched/fair: Fix enqueue_task_fair warning some more

[Dietmar Eggemann]

On 11/05/2020 14:12, Vincent Guittot wrote:
> On Mon, 11 May 2020 at 12:39, Dietmar Eggemann <dietmar.eggemann@xxxxxxx> wrote:
>>
>> On 11/05/2020 11:36, Vincent Guittot wrote:
>>> On Mon, 11 May 2020 at 10:40, Dietmar Eggemann <dietmar.eggemann@xxxxxxx> wrote:
>>>>
>>>> On 08/05/2020 19:02, Tao Zhou wrote:
>>>>> On Fri, May 08, 2020 at 05:27:44PM +0200, Vincent Guittot wrote:
>>>>>> On Fri, 8 May 2020 at 17:12, Tao Zhou <zohooouoto@xxxxxxxxxxx> wrote:
>>>>>>>
>>>>>>> Hi Phil,
>>>>>>>
>>>>>>> On Thu, May 07, 2020 at 04:36:12PM -0400, Phil Auld wrote:
>>>>>>>> sched/fair: Fix enqueue_task_fair warning some more
>>
>> [...]
>>
>>>> I'm not 100% sure if this is exactly what Tao pointed out here but I
>>>> also had difficulties understanding understanding how this patch works:
>>>>
>>>>                        p.se
>>>>                         |
>>>>       __________________|
>>>>       |
>>>>       V
>>>>      cfs_c -> tg_c ->  se_c (se->on_rq = 1)
>>>>                         |
>>>>       __________________|
>>>>       |
>>>>       v
>>>>      cfs_b -> tg_b ->  se_b
>>>>                         |
>>>>       __________________|
>>>>       |
>>>>       V
>>>>      cfs_a -> tg_a ->  se_a
>>>>                         |
>>>>       __________________|
>>>>       |
>>>>       V
>>>>      cfs_r -> tg_r
>>>>       |
>>>>       V
>>>>       rq
>>>>
>>>
>>> In your example, which cfs_ rq has been throttled ? cfs_a ?
>>
>> Yes, cfs_a. 0xffffa085e48ce000 in Phil's trace.
>>
>>>
>>>> (1) The incomplete update happens with cfs_c at the end of
>>>>     enqueue_entity() in the first loop because of 'if ( .... ||
>>>>     cfs_bandwidth_used())' (cfs_b->on_list=0 since cfs_a is throttled)
>>>
>>> so cfs_c is added with the 1st loop
>>
>> Yes.
>>
>>>> (2) se_c breaks out of the first loop (se_c->on_rq = 1)
>>>>
>>>> (3) With the patch cfs_b is added back to the list.
>>>>     But only because cfs_a->on_list=1.
>>>
>>> hmm I don't understand the link between cfs_b been added and cfs_a->on_list=1
>>
>> cfs_b, 0xffffa085e48ce000 is the one which is now added in the 2. loop.
>>
>> Isn't the link between cfs_b and cfs_a the first if condition in
> 
> on_list is only there to say if the cfs_rq is already in the list but
> there is not dependency with the child

Yes, I agree. But coming back to what the patch does in the example:

W/ the patch, list_add_leaf_cfs_rq() is now called for cfs_b and since
cfs_b->tg->parent->cfs_a and cfs_a->on_list=1 the 'branch is now
connected' which means 'rq->tmp_alone_branch = &rq->leaf_cfs_rq_list'.

I.e. assert_list_leaf_cfs_rq() at the end of enqueue_task_fair() is not
barfing anymore.

W/o the patch, list_add_leaf_cfs_rq() called w/ cfs_c left the 'branch
open', it's not called on cfs_b and since cfs_a->on_list=1, the 3rd
for_each_sched_entity() in enqueue_task_fair() doesn't 'connect the
branch' so the assert fires.

What I don't immediately see is how can cfs_a be throttled (which causes
cfs_b -> cfs_c being a throttled hierarchy) and be on the list
(cfs_a->on_list=1) at the same time.

So the only thing how this could happen is when there was a task enqueue
in a parallel cfs_b' (another child of cfs_a) sub hierarchy just before
the example.

>> list_add_leaf_cfs_rq():
>>
>>   if (cfs_rq->tg->parent &&
>>       cfs_rq->tg->parent->cfs_rq[cpu]->on_list)
>>
>> to 'connect the branch' or not (default, returning false case)?
>>
> 
> In your example above if the parent is already on the list then we
> know where to insert the child.

True, we go the 2nd if() condition in list_add_leaf_cfs_rq().

>>> cfs_b is added with 2nd loop because its throttle_count > 0 due to
>>> cfs_a been throttled (purpose of this patch)
>>>
>>>>
>>>> But since cfs_a is throttled it should be cfs_a->on_list=0 as well.
>>>
>>> So 2nd loop breaks because cfs_a is throttled
>>
>> Yes.
>>
>>> The 3rd loop will add cfs_a
>>
>> Yes, but in the example, cfs_a->on_list=1, so we bail out of
>> list_add_leaf_cfs_rq() early.
> 
> Because the cfs_rq is on the list already so we don't have to add it

Yes.

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