Re: [PATCH] de_thread: Move notify_count write under lock

[Oleg Nesterov]

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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