Re: [PATCH 08/14] hrtimer: Allow hrtimer::function() to free the timer

[Oleg Nesterov <oleg@redhat.com>]

Hi Kirill,

On 06/10, Kirill Tkhai wrote:
>
> В Вт, 09/06/2015 в 23:33 +0200, Oleg Nesterov пишет:
> >
> > 	hrtimer_active(timer)
> > 	{
> >
> > 		do {
> > 			base = READ_ONCE(timer->base->cpu_base);
> > 			seq = read_seqcount_begin(&cpu_base->seq);
> >
> > 			if (timer->state & ENQUEUED ||
> > 			    base->running == timer)
> > 				return true;
> >
> > 		} while (read_seqcount_retry(&cpu_base->seq, seq) ||
> > 			 base != READ_ONCE(timer->base->cpu_base));
> >
> > 		return false;
> > 	}
> >
> > And we need to avoid the races with 2 transitions in __run_hrtimer().
> >
> > The first race is trivial, we change __run_hrtimer() to do
> >
> > 	write_seqcount_begin(cpu_base->seq);
> > 	cpu_base->running = timer;
> > 	__remove_hrtimer(timer);	// clears ENQUEUED
> > 	write_seqcount_end(cpu_base->seq);
>
> We use seqcount, because we are afraid that hrtimer_active() may miss
> timer->state or cpu_base->running, when we are clearing it.

Yes,

> If we use two pairs of write_seqcount_{begin,end} in __run_hrtimer(),
> we may protect only the places where we do that:
>
> 	cpu_base->running = timer;
> 	write_seqcount_begin(cpu_base->seq);
> 	__remove_hrtimer(timer);	// clears ENQUEUED
> 	write_seqcount_end(cpu_base->seq);
>
> 	....
>
> 	timer->state |= HRTIMER_STATE_ENQUEUED;
> 	write_seqcount_begin(cpu_base->seq);
> 	base->running = NULL;
> 	write_seqcount_end(cpu_base->seq);

Afaics, no. Afaics, the following code is correct:

	seqcount_t LOCK;
	bool X = true, Y = false;

	void read(void)
	{
		bool x, y;

		do {
			seq = read_seqcount_begin(&LOCK);

			x = X; y = Y;

		} while (read_seqcount_retry(&LOCK, seq));

		BUG_ON(!x && !y);
	}

	void write(void)
	{
		Y = true;

		write_seqcount_begin(LOCK);
		write_seqcount_end(LOCK);

		X = false;
	}

If we rely on the "locking" semantics of seqcount_t, this doesn't really
differ from

	spinlock_t LOCK;
	bool X = true, Y = false;

	void read(void)
	{
		bool x, y;

		spin_lock(LOCK);
		x = X; y = Y;
		spin_unlock(LOCK);

		BUG_ON(!x && !y);
	}

	void write(void)
	{
		Y = true;

		spin_lock(LOCK);
		spin_unlock(LOCK);

		X = false;
	}

If "read" takes the lock before "write", it must see X == true.

Otherwise "read" should see all memory changes done before or
inside the "write" critical section, so it must see Y == true.

No?

Oleg.