Re: [patch 3/21] x86, bts: wait until traced task has beenscheduled out

[Oleg Nesterov]

On 04/01, Metzger, Markus T wrote:
>
> >-----Original Message-----
> >From: Oleg Nesterov [mailto:oleg@xxxxxxxxxx]
> >Sent: Wednesday, April 01, 2009 2:17 AM
> >To: Metzger, Markus T
>
> >> +static void wait_to_unschedule(struct task_struct *task)
> >> +{
> >> +	unsigned long nvcsw;
> >> +	unsigned long nivcsw;
> >> +
> >> +	if (!task)
> >> +		return;
> >> +
> >> +	if (task == current)
> >> +		return;
> >> +
> >> +	nvcsw  = task->nvcsw;
> >> +	nivcsw = task->nivcsw;
> >> +	for (;;) {
> >> +		if (!task_is_running(task))
> >> +			break;
> >> +		/*
> >> +		 * The switch count is incremented before the actual
> >> +		 * context switch. We thus wait for two switches to be
> >> +		 * sure at least one completed.
> >> +		 */
> >> +		if ((task->nvcsw - nvcsw) > 1)
> >> +			break;
> >> +		if ((task->nivcsw - nivcsw) > 1)
> >> +			break;
> >> +
> >> +		schedule();
> >
> >schedule() is a nop here. We can wait unpredictably long...
>
> Hmmm, As far as I understand the code, rt-workqueues use a higher sched_class
> and can thus not be preempted by normal threads. Non-rt workqueues
> use the fair_sched_class. And schedule_work() uses a non-rt workqueue.

I was unclear, sorry.

I meant, in this case

	while (!CONDITION)
		schedule();

is not better compared to

	while (!CONDITION)
		; /* do nothing */

(OK, schedule() is better without CONFIG_PREEMPT, but this doesn't matter).
wait_to_unschedule() just spins waiting for ->nXvcsw, this is not optimal.

And another problem, we can wait unpredictably long, because

> In practice, task is ptraced. It is either stopped or exiting.
> I don't expect to loop very often.

No. The task _was_ ptraced when we called (say) ptrace_detach(). But when
work->func() runs, the tracee is not traced, it is running (not necessary
of course, the tracer _can_ leave it in TASK_STOPPED).

Now, again, suppose that this task does "for (;;) ;" in user-space.
If CPU is "free", it can spin "forever" without re-scheduling. Yes sure,
this case is not likely in practice, but still.

Oleg.

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