Hi Mandeep,

On 01/11, Mandeep Singh Baines wrote:

quoted

quoted

#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.