Re: Q: cgroup: Questions about possible issues in cgroup locking

[Oleg Nesterov <oleg-H+wXaHxf7aLQT0dZR+AlfA@public.gmane.org>]

On 01/12, Mandeep Singh Baines wrote:
>
> Oleg Nesterov (oleg-H+wXaHxf7aLQT0dZR+AlfA@public.gmane.org) wrote:
> >
> > Still can't understand... Lets look at this trivial example again.
> >
> > We start from the main thread M, it is ->group_leader. There is
> > another thread T in this thread group. We are doing
> >
> > 	OLD = M;
> >
> > 	t = M;
> > 	do {
> > 		do_smth(t);
> > 	}
> > 	while (t->group_leader == OLD && ((t = next_thread(t)) != M);
> >
> > The first iteration does do_smth(M).
> >
> > T calls de_thread() and, in particular, it does M->group_leader = T
> > (see "leader->group_leader = tsk" in de_thread).
> >
> > after that t->group_leader == OLD fails. t == M, its group_leader == T.
> > do_smth(T) won't be called.
> >
> > No?
> >
>
> I think we can handle this by removing the assignment. So in de_thread():
>
> -		leader->group_leader = tsk;

Ah, so that was you plan. I was confused by the 3rd argument, why
it is needed?

Yes, I thought about this too. Suppose we remove this assignment,
then we can simply do

	#define while_each_thread(g, t) \
		while (t->group_leader == g->group_leader && (t = next_thread(t)) != g)

with the same effect. (to remind, currently I ignore the barriers/etc).

But this can _only_ help if we start at the group leader!

May be we should enforce this rule (for the lockless case), I dunno...
In that case I'd prefer to add the new while_each_thread_rcu() helper.
But! in this case we do not need to change de_thread(), we can simply do

	#define while_each_thread_rcu(t) \
		while (({ t = next_thread(t); !thread_group_leader(t); }))

The definition above was one of the possibilities I considered, but
I wasn't able to convince myself this is the best option.

See? Or do you think I missed something?

Just in case... note that while_each_thread_rcu() doesn't use 'g'
at all. May be it makes sense to keep the old "t != g &&", but this
is minor.

Oleg.