Re: [PATCH] sched/fair: fix pick_eevdf to always find the correct se

[Benjamin Segall]

Abel Wu <wuyun.abel@xxxxxxxxxxxxx> writes:

> On 10/12/23 5:01 AM, Benjamin Segall Wrote:
>> Abel Wu <wuyun.abel@xxxxxxxxxxxxx> writes:
>> 
>>> On 9/30/23 8:09 AM, Benjamin Segall Wrote:
>>>> +	/*
>>>> +	 * Now best_left and all of its children are eligible, and we are just
>>>> +	 * looking for deadline == min_deadline
>>>> +	 */
>>>> +	node = &best_left->run_node;
>>>> +	while (node) {
>>>> +		struct sched_entity *se = __node_2_se(node);
>>>> +
>>>> +		/* min_deadline is the current node */
>>>> +		if (se->deadline == se->min_deadline)
>>>> +			return se;
>>>
>>> IMHO it would be better tiebreak on vruntime by moving this hunk to ..
>>>
>>>> +
>>>> +		/* min_deadline is in the left branch */
>>>>    		if (node->rb_left &&
>>>>    		    __node_2_se(node->rb_left)->min_deadline == se->min_deadline) {
>>>>    			node = node->rb_left;
>>>>    			continue;
>>>>    		}
>>>
>>> .. here, thoughts?
>> Yeah, that should work and be better on the tiebreak (and my test code
>> agrees). There's an argument that the tiebreak will never really come up
>> and it's better to avoid the potential one extra cache line from
>> "__node_2_se(node->rb_left)->min_deadline" though.
>
> I see. Then probably do the same thing in the first loop?
>

We effectively do that already sorta by accident almost always -
computing best and best_left via deadline_gt rather than gte prioritizes
earlier elements, which always have a better vruntime.

Then when we do the best_left->min_deadline vs best->deadline
computation, we prioritize best_left, which is the one case it can be
wrong, we'd need an additional
"if (se->min_deadline == best->deadline &&
(s64)(se->vruntime - best->vruntime) > 0) return best;" check at the end
of the second loop.

(Though again I don't know how much this sort of never-going-to-happen
slight fairness improvement is worth compared to the extra bit of
overhead)