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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