Re: [patch V5 16/26] signal: Replace resched_timer logic

[Thomas Gleixner <tglx@linutronix.de>]

On Tue, Oct 29 2024 at 17:34, Frederic Weisbecker wrote:

> Le Tue, Oct 29, 2024 at 05:22:17PM +0100, Thomas Gleixner a écrit :
>> On Tue, Oct 29 2024 at 16:56, Frederic Weisbecker wrote:
>> >> @@ -568,10 +568,10 @@ static void collect_signal(int sig, stru
>> >>  		list_del_init(&first->list);
>> >>  		copy_siginfo(info, &first->info);
>> >>  
>> >> -		*resched_timer = (first->flags & SIGQUEUE_PREALLOC) &&
>> >> -				 (info->si_code == SI_TIMER);
>> >> -
>> >> -		__sigqueue_free(first);
>> >> +		if (unlikely((first->flags & SIGQUEUE_PREALLOC) && (info->si_code == SI_TIMER)))
>> >> +			*timer_sigq = first;
>> >> +		else
>> >> +			__sigqueue_free(first);
>> >
>> > So this isn't calling __sigqueue_free() unconditionally anymore. What if
>> > the timer has been freed already, what is going to free the sigqueue?
>> 
>> __sigqueue_free() does not free timers marked with SIGQUEUE_PREALLOC.
>> 
>> sigqueue_free() takes care of that, which is invoked from
>> posixtimer_free_timer(). It clears SIGQUEUE_PREALLOC and if it is queued
>> it lets it pending and delivery will free it.
>
> But the delivery freeing used to be done with the __sigqueue_free()
> above, which doesn't happen anymore, right?

It still happens because SIGQUEUE_PREALLOC is cleared in sigqueue_free()

__sigqueue_free() has
       if (q->flags & PREALLOC)
       	     return;

So the old code called __sigqueue_free() unconditionally which just
returned. But now we have a condition to that effect already, so why
call into __sigqueue_free() for nothing?

Let me add a comment.

Thanks,

        tglx