Re: [RFD/RFC PATCH 5/8] sched: Add proxy execution

[luca abeni]

On Thu, 11 Oct 2018 14:53:25 +0200
Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:

[...]
> > > > +	if (rq->curr != rq->idle) {
> > > > +		rq->proxy = rq->idle;
> > > > +		set_tsk_need_resched(rq->idle);
> > > > +		/*
> > > > +		 * XXX [juril] don't we still need to migrate
> > > > @next to
> > > > +		 * @owner's CPU?
> > > > +		 */
> > > > +		return rq->idle;
> > > > +	}  
> > > 
> > > If I understand well, this code ends up migrating the task only
> > > if the CPU was previously idle? (scheduling the idle task if the
> > > CPU was not previously idle)
> > > 
> > > Out of curiosity (I admit this is my ignorance), why is this
> > > needed? If I understand well, after scheduling the idle task the
> > > scheduler will be invoked again (because of the
> > > set_tsk_need_resched(rq->idle)) but I do not understand why it is
> > > not possible to migrate task "p" immediately (I would just check
> > > "rq->curr != p", to avoid migrating the currently scheduled
> > > task).  
[...]
> I think it was the safe and simple choice; note that we're not
> migrating just a single @p, but a whole chain of @p.

Ah, that's the point I was missing... Thanks for explaining, now
everything looks more clear!


But... Here is my next dumb question: once the tasks are migrated to
the other runqueue, what prevents the scheduler from migrating them
back? In particular, task p: if it is (for example) a fixed priority
task an is on this runqueue, it is probably because the FP invariant
wants this... So, the push mechanism might end up migrating p back to
this runqueue soon... No?

Another doubt: if I understand well, when a task p "blocks" on a mutex
the proxy mechanism migrates it (and the whole chain of blocked tasks)
to the owner's core... Right?
Now, I understand why this is simpler to implement, but from the
schedulability point of view shouldn't we migrate the owner to p's core
instead?


			Thanks,
				Luca

> rq->curr must
> not be any of the possible @p's. rq->idle, is per definition not one
> of the @p's.
> 
> Does that make sense?