Re: [PATCH v2 10/10] sched/eevdf: Move to a single runqueue

[Peter Zijlstra]

On Thu, May 21, 2026 at 03:29:01PM +0200, Peter Zijlstra wrote:
> On Thu, May 21, 2026 at 02:13:48PM +0200, Vincent Guittot wrote:
> 
> > > Would it not be simpler to just move the update_entity_lag() call up a
> > > bit, like so?
> > >
> > > ---
> > > --- a/kernel/sched/fair.c
> > > +++ b/kernel/sched/fair.c
> > > @@ -7999,6 +7999,9 @@ static bool __dequeue_task(struct rq *rq
> > >
> > >         clear_buddies(cfs_rq, se);
> > >
> > > +       update_curr(cfs_rq);
> > 
> > I agree it's simpler although we will call update_curr twice for one
> > level, but the 2nd call should be nop because of delta_exec being null
> > 
> > Prateek proposed update_curr(task_cfs_rq(p)). Using task_cfs_rq(p)
> > will ensure that we keep the same ordering as for_each_sched_entity
> 
> Given:
> 
>     R
>     |
>     G
>     |
>     t
> 
> Then task_cfs_rq() will be G's cfs_rq, while cfs_rq is R's cfs_rq.
> 
> Since all the actual running happens inside R, this is what is required
> by update_entity_lag().
> 
> Doing update_curr(task_cfs_rq()) here doesn't make sense.
> 
> I'm not sure I see a way in which running them out of order hurts
> anything.

Bah, I'm so full of fail. So update_curr() takes ->h_curr, which for R
would be G's se, not t. So yeah, Prateek is right and I should stop
trying to do more than one thing at a time :-(