Re: [PATCH 3/3] Consolidate send_sigqueue and send_group_sigqueue

[Thomas Gleixner]

On Fri, 22 Feb 2008, Oleg Nesterov wrote:
> > -
> > -	if (unlikely(!list_empty(&q->list))) {
> > -		/*
> > -		 * If an SI_TIMER entry is already queue just increment
> > -		 * the overrun count.
> > -		 */
> > -		BUG_ON(q->info.si_code != SI_TIMER);
> > -		q->info.si_overrun++;
> > -		goto out;
> > -	}
> > -	/* Short-circuit ignored signals.  */
> > -	if (sig_ignored(p, sig)) {
> > -		ret = 1;
> > -		goto out;
> > -	}

> >  send_group_sigqueue(int sig, struct sigqueue *q, struct task_struct *p)

> > -	/* Short-circuit ignored signals.  */
> > -	if (sig_ignored(p, sig)) {
> > -		ret = 1;
> > -		goto out;
> > -	}
> > -
> > -	if (unlikely(!list_empty(&q->list))) {
> > -		/*
> > -		 * If an SI_TIMER entry is already queue just increment
> > -		 * the overrun count.  Other uses should not try to
> > -		 * send the signal multiple times.
> > -		 */
> > -		BUG_ON(q->info.si_code != SI_TIMER);
> > -		q->info.si_overrun++;
> > -		goto out;
> > -	} 
> 
> Personally, I think this change is very good. But send_sigqueue() and
> send_group_sigqueue() have a very subtle difference which I was never
> able to understand.
> 
> Let's suppose that sigqueue is already queued, and the signal is ignored
> (the latter means we should re-schedule cpu timer or handle overrruns).
> In that case send_sigqueue() returns 0, but send_group_sigqueue() returns 1.
> 
> I think this is not the problem (in fact, I think this patch makes the
> behaviour more correct), but I hope Thomas can take a look and confirm.

It should not change anything. We should never have a signal enqueued
when it's ignored anyway.

Roland, any insight why this is different aside of a copy and paste
error ?

Thanks,

	tglx
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