Re: [RFC][PATCH 1/5] signals: Always place SIGCONT and SIGSTOP on'shared_pending'

From: Matt Fleming
Date: Tue Apr 05 2011 - 16:50:38 EST


On Tue, 5 Apr 2011 22:19:58 +0200
Oleg Nesterov <oleg@xxxxxxxxxx> wrote:

> Hi Matt,
>
> I'll try to study this series, but not before Friday, sorry.

No problem!

> Only one thing,
>
> On 04/05, Matt Fleming wrote:
> >
> > Because SIGCONT and SIGSTOP affect an entire thread group,
>
> Yes, the effect is global, but
>
> > we can
> > place them on the 'shared_pending' queue.
>
> I don't think we can.
>
> - pending = group ? &t->signal->shared_pending : &t->pending;
> > + /*
> > + * We always enqueue SIGSTOP or SIGCONT signals on the
> > shared
> > + * queue. This means that a SIGSTOP or SIGCONT signal
> > _cannot_
> > + * be present on a thread's private pending queue.
> > + *
> > + * This makes prepare_signal() more optimal as we do not
> > have
> > + * to remove signals from each thread's pending queue and
> > so
> > + * can avoid iterating over all threads in the thread group
> > + * (and therefore avoid the locking that would be
> > necessary to
> > + * do that safely).
> > + */
> > + if (group || sig_kernel_stop(sig) || sig == SIGCONT)
> > + pending = &t->signal->shared_pending;
> > + else
> > + pending = &t->pending;
>
> How so? Suppose the process has a handler for SIGCONT. Suppose this
> process is not stopped. tkill(SIGCONT) should deliver the signal to
> the right thread.

D'oh, yes. I think I got confused here. You're right, this won't work.

> SIGSTOP can't have the handler, still we shouldn't place it on the
> shared list, debuggers won't be happy.

Urgh, debuggers actually peek at shared_pending and pending?

> Also. This code was changed very much, please do these changes on
> top of
> http://git.kernel.org/?p=linux/kernel/git/tj/misc.git;a=shortlog;h=refs/heads/ptrace

My patches are already based on that tree.

--
Matt Fleming, Intel Open Source Technology Center
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