Re: [patch V2 3/3] signal: Allow tasks to cache one sigqueue struct

[Oleg Nesterov]

On 03/12, Thomas Gleixner wrote:
>
> On Fri, Mar 12 2021 at 20:26, Thomas Gleixner wrote:
> > On Fri, Mar 12 2021 at 17:11, Oleg Nesterov wrote:
> >> On 03/11, Thomas Gleixner wrote:
> >>>
> >>> @@ -456,7 +460,12 @@ static void __sigqueue_free(struct sigqu
> >>>  		return;
> >>>  	if (atomic_dec_and_test(&q->user->sigpending))
> >>>  		free_uid(q->user);
> >>> -	kmem_cache_free(sigqueue_cachep, q);
> >>> +
> >>> +	/* Cache one sigqueue per task */
> >>> +	if (!current->sigqueue_cache)
> >>> +		current->sigqueue_cache = q;
> >>> +	else
> >>> +		kmem_cache_free(sigqueue_cachep, q);
> >>>  }
> >>
> >> This doesn't look right, note that __exit_signal() does
> >> flush_sigqueue(&sig->shared_pending) at the end, after exit_task_sighand()
> >> was already called.
> >>
> >> I'd suggest to not add the new exit_task_sighand() helper and simply free
> >> current->sigqueue_cache at the end of __exit_signal().
> >
> > Ooops. Thanks for spotting this!
>
> Hrm.
>
> The task which is released is obviously not current, so even if there
> are still sigqueues in shared_pending then they wont end up in the
> released tasks sigqueue_cache. They can only ever end up in
> current->sigqueue_cache.

The task which is released can be "current" if this task autoreaps.
For example, if its parent ignores SIGCHLD. In this case exit_notify()
does release_task(current).

> But that brings my memory back why I had cmpxchg() in the original
> version. This code runs without current->sighand->siglock held.

Yes, I was wrong too. This code runs without exiting_task->sighand->siglock
and this is fine in that it can not race with send_signal(exiting_task),
but somehow I missed the fact that it always populates current->sigqueue_cache,
not exiting_task->sigqueue_cache.

Oleg.