Re: [PATCH 2/3] sched/fair: Make cfs_rq::idle the sum of ancestor tg::idle

[Kayra Cizmeci]

On Fri, Nov 06, 2026 at 09:31:06AM +0100, Peter Zijlstra wrote:
> On Thu, Nov 05, 2026 at 10:20:00PM +0300, Kayra Cizmeci wrote:
> > On Thu, Nov 05, 2026 at 05:55:27PM +0100, Peter Zijlstra wrote:
> > > Rather than keeping cfs_rq::idle as a copy of tg::idle, make it the sum of
> > > all ancestor tg::idle. This makes it trivial to tell if the current cgroup
> > > is marked idle somewhere along its ancestral path.
> > > 
> > > This makes se_id_idle() behave differently. Notably is will now be true
> > > when there is a idle cgroup anywhere in the ancestor path. This
> > > addresses an issue where preemption of cgroup::idle was overlooked when
> > > flattening the runqueue.
> > 
> > How? The diff doesn't touches se_is_idle(). se_is_idle() just looks to
> > the policy for tasks:
> > 
> > static int se_is_idle(struct sched_entity *se)
> > {
> > 	if (entity_is_task(se))
> > 		return task_has_idle_policy(task_of(se));
> > 	return cfs_rq_is_idle(group_cfs_rq(se));
> > }
> 
> > Or am I getting something wrong?
> 
> cfs_rq_is_idle() := cfs_rq->idle > 0
> 
> and cfs_rq->idle is now getting incremented for the whole subtree. So if
> any ancestor is idle, this is now true.


Yeah, it is for group path. But we call se_is_idle for tasks in
wakeup_preempt_fair(). Not groups, no?