Messages in this thread | | | Date | Thu, 12 Jan 2012 18:07:28 +0100 | From | Oleg Nesterov <> | Subject | Re: Q: cgroup: Questions about possible issues in cgroup locking |
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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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