On 01/12, Mandeep Singh Baines wrote:
>
> Oleg Nesterov (oleg@xxxxxxxxxx) 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.
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