Messages in this thread | | | Date | Thu, 5 Feb 2015 14:38:29 +0100 | From | Oleg Nesterov <> | Subject | Re: [PATCH] de_thread: Move notify_count write under lock |
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On 02/05, Kirill Tkhai wrote: > > The write operation may be reordered with the setting of group_exit_task. > If so, this fires in exit_notify().
How?
OK, yes, "sig->notify_count = -1" can be reordered with the last unlock, but we do not care?
group_exit_task + notify_count is only checked under the same lock, and "notify_count = -1" can't happen until de_thread() sees it is zero.
Could you explain why this is bad in more details?
> --- a/fs/exec.c > +++ b/fs/exec.c > @@ -920,10 +920,16 @@ static int de_thread(struct task_struct *tsk) > if (!thread_group_leader(tsk)) { > struct task_struct *leader = tsk->group_leader; > > - sig->notify_count = -1; /* for exit_notify() */ > for (;;) { > threadgroup_change_begin(tsk); > write_lock_irq(&tasklist_lock); > + /* > + * We could set it once outside the for() cycle, but > + * this requires to use SMP barriers there and in > + * exit_notify(), because the write operation may > + * be reordered with the setting of group_exit_task. > + */ > + sig->notify_count = -1; /* for exit_notify() */ > if (likely(leader->exit_state)) > break; > __set_current_state(TASK_KILLABLE);
Perhaps something like this makes sense anyway to make the code more clear, but in this case I'd suggest to set ->notify_count after we check ->exit_state. And without the (afaics!) misleading comment...
Or I missed something?
Oleg.
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