Re: [RFC][PATCH 5/5] sched: Reduce ttwu rq->lock contention

[Oleg Nesterov]

On 12/17, Oleg Nesterov wrote:
>
> On 12/16, Peter Zijlstra wrote:
> >
> > +	if (p->se.on_rq && ttwu_force(p, state, wake_flags))
> > +		return 1;
>
> 	----- WINDOW -----
>
> > +	for (;;) {
> > +		unsigned int task_state = p->state;
> > +
> > +		if (!(task_state & state))
> > +			goto out;
> > +
> > +		load = task_contributes_to_load(p);
> > +
> > +		if (cmpxchg(&p->state, task_state, TASK_WAKING) == task_state)
> > +			break;
>
> Suppose that we have a task T sleeping in TASK_INTERRUPTIBLE state,
> and this cpu does try_to_wake_up(TASK_INTERRUPTIBLE). on_rq == false.
> try_to_wake_up() starts the "for (;;)" loop.
>
> However, in the WINDOW above, it is possible that somebody else wakes
> it up, and then this task changes its state to TASK_INTERRUPTIBLE again.
>
> Then we set ->state = TASK_WAKING, but this (still running) T restores
> TASK_RUNNING after us.

Even simpler. This can race with, say, __migrate_task() which does
deactivate_task + activate_task and temporary clears on_rq. Although
this is simple to fix, I think.



Also. Afaics, without rq->lock, we can't trust "while (p->oncpu)", at
least we need rmb() after that.

Interestingly, I can't really understand the current meaning of smp_wmb()
in finish_lock_switch(). Do you know what exactly is buys? In any case,
task_running() (or its callers) do not have the corresponding rmb().
Say, currently try_to_wake_up()->task_waking() can miss all changes
starting from prepare_lock_switch(). Hopefully this is OK, but I am
confused ;)

Oleg.

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