On Wed, 30 Jan 2019 at 15:27, Vincent Guittot <vincent.guit...@linaro.org> wrote: > > On Wed, 30 Jan 2019 at 15:01, Peter Zijlstra <pet...@infradead.org> wrote: > > > > On Wed, Jan 30, 2019 at 03:01:04PM +0100, Peter Zijlstra wrote: > > > --- a/kernel/sched/fair.c > > > +++ b/kernel/sched/fair.c > > > @@ -282,13 +282,15 @@ static inline struct cfs_rq *group_cfs_r > > > return grp->my_q; > > > } > > > > > > -static inline void list_add_leaf_cfs_rq(struct cfs_rq *cfs_rq) > > > +static inline bool list_add_leaf_cfs_rq(struct cfs_rq *cfs_rq) > > > { > > > struct rq *rq = rq_of(cfs_rq); > > > int cpu = cpu_of(rq); > > > > > > if (cfs_rq->on_list) > > > - return; > > > + return rq->tmp_alone_branch == &rq->leaf_cfs_rq_list; > > > > And I'm almost certain that can be: return true, but got my brain in a > > twist. > > Yes this can return true > > If cfs_rq->on_list) then a child not already on the list used the path : > > if (cfs_rq->tg->parent && > cfs_rq->tg->parent->cfs_rq[cpu]->on_list) { > > which does rq->tmp_alone_branch = &rq->leaf_cfs_rq_list;
In fact tests show that we must keep: return rq->tmp_alone_branch == &rq->leaf_cfs_rq_list; Because the 1st sched_entity that will be used in the newly added for_each_sched_entity loop, can be the sched_entityof the cfs_rq that we just added in the list so cfs_rq->on_list == 1 but we must continue to add parent Apart from that, tests are ok > > > > > > + > > > + cfs_rq->on_list = 1; > > > > > > /* > > > * Ensure we either appear before our parent (if already