Re: [RFC -v6 PATCH 4/8] sched: Add yield_to(task, preempt) functionality

[Rik van Riel <riel@redhat.com>]

On 01/24/2011 01:12 PM, Peter Zijlstra wrote:
> On Thu, 2011-01-20 at 16:34 -0500, Rik van Riel wrote:
>> From: Mike Galbraith<efault@gmx.de>
>>
>> Currently only implemented for fair class tasks.
>>
>> Add a yield_to_task method() to the fair scheduling class. allowing the
>> caller of yield_to() to accelerate another thread in it's thread group,
>> task group.
>>
>> Implemented via a scheduler hint, using cfs_rq->next to encourage the
>> target being selected.  We can rely on pick_next_entity to keep things
>> fair, so noone can accelerate a thread that has already used its fair
>> share of CPU time.
>>
>> This also means callers should only call yield_to when they really
>> mean it.  Calling it too often can result in the scheduler just
>> ignoring the hint.
>>
>> Signed-off-by: Rik van Riel<riel@redhat.com>
>> Signed-off-by: Marcelo Tosatti<mtosatti@redhat.com>
>> Signed-off-by: Mike Galbraith<efault@gmx.de>
>
> Patch 5 wants to be merged back in here I think..

Agreed, but I wanted Mike's comments first  :)


>> +/**
>> + * yield_to - yield the current processor to another thread in
>> + * your thread group, or accelerate that thread toward the
>> + * processor it's on.
>> + *
>> + * It's the caller's job to ensure that the target task struct
>> + * can't go away on us before we can do any checks.
>> + *
>> + * Returns true if we indeed boosted the target task.
>> + */
>> +bool __sched yield_to(struct task_struct *p, bool preempt)
>> +{
>> +	struct task_struct *curr = current;
>> +	struct rq *rq, *p_rq;
>> +	unsigned long flags;
>> +	bool yielded = 0;
>> +
>> +	local_irq_save(flags);
>> +	rq = this_rq();
>> +
>> +again:
>> +	p_rq = task_rq(p);
>> +	double_rq_lock(rq, p_rq);
>> +	while (task_rq(p) != p_rq) {
>> +		double_rq_unlock(rq, p_rq);
>> +		goto again;
>> +	}
>> +
>> +	if (!curr->sched_class->yield_to_task)
>> +		goto out;
>> +
>> +	if (curr->sched_class != p->sched_class)
>> +		goto out;
>> +
>> +	if (task_running(p_rq, p) || p->state)
>> +		goto out;
>> +
>> +	if (!same_thread_group(p, curr))
>> +		goto out;
>> +
>> +#ifdef CONFIG_FAIR_GROUP_SCHED
>> +	if (task_group(p) != task_group(curr))
>> +		goto out;
>> +#endif
>> +
>> +	yielded = curr->sched_class->yield_to_task(rq, p, preempt);
>> +
>> +out:
>> +	double_rq_unlock(rq, p_rq);
>> +	local_irq_restore(flags);
>> +
>> +	if (yielded)
>> +		yield();
>
> Calling yield() here is funny, you just had all the locks to actually do
> it..

This is us giving up the CPU, which requires not holding locks.

A different thing than us giving the CPU away to someone else.

>> +static bool yield_to_task_fair(struct rq *rq, struct task_struct *p, bool preempt)
>> +{
>> +	struct sched_entity *se =&p->se;
>> +
>> +	if (!se->on_rq)
>> +		return false;
>> +
>> +	/* Tell the scheduler that we'd really like pse to run next. */
>> +	set_next_buddy(se);
>> +
>> +	/* Make p's CPU reschedule; pick_next_entity takes care of fairness. */
>> +	if (preempt)
>> +		resched_task(rq->curr);
>> +
>> +	return true;
>> +}
>
> So here we set ->next, we could be ->last, and after this we'll set
> ->yield to curr by calling yield().
>
> So if you do this cyclically I can see ->yield == {->next,->last}
> happening.

That would only happen if we called yield_to with ourselves
as the argument!

There is no caller in the tree that does that - task p is
another task, not ourselves.