Hi Mandeep,
On 01/11, Mandeep Singh Baines wrote:
> > >
> > > #define while_each_thread(g, t, o) \
> > > while (t->group_leader == o && (t = next_thread(t)) != g)
> > >
> > > Where o should have the value of g->group_leader.
> >
> > I don't understand how this helps... and how this can work even
> > ignoring the barriers.
> >
> > OK, we have the main thream M and the sub-thread T, we are doing
> >
> > do {
> > do_something(t);
> > } while_each_thread(M, t, M);
> >
> > why we can't miss T if it does exec?
> >
>
> So for:
>
> struct task *M; /* assuming this is passed in to us */
> struct task *L = M->group_leader;
L == M
> do {
> do_something(T);
> } while_each_thread(M, T, L);
>
> Here is my thinking.
>
> If some thread K does exec, you won't miss it because:
>
> 1) Ignoring the group_leader check, you'll visit K just by following
> next_thread(). That's the case today and is what you except
> when iterating over an rcu_list.
> 2) (t->group_leader == o) will fail iff t is the exec thread.
> Since we test t->group_leader before re-assigning it (t=next_thread()),
> the test will fail only after visiting the exec thread. So you'll
> visit the exec thread and then terminate the loop.
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?
Oleg.
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