Hi Petr,
On 06/15, Petr Mladek wrote:
>
> I am sorry for the late reply. I wanted to think more before answering
> all the mails.
Don't worry I am always late ;)
> On Mon 2015-06-08 23:13:36, Oleg Nesterov wrote:
> > >
> > > Hmm, the helper would have a strange semantic. You need to take
> > > sighand->siglock, dequeue the signal (SIGSTOP), and call
> > > __set_current_state(TASK_STOPPED) before you release the lock.
> > > But what would happen if the dequeued signal is _not_ SIGSTOP?
> >
> > Perhaps I missed your point, but no. If you want to handle SIGSTOP
> > you can do
> >
>
> I think that we need to add:
>
> spin_lock_irq(&sighand->siglock);
>
> > signr = kthread_signal_dequeue();
> > switch (signr) {
> > case SIGSTOP:
> > something_else();
> > kthread_do_signal_stop();
> > ...
> > }
>
> And if we want to avoid any race, kthread_do_signal_stop() should look like:
>
> void kthread_do_signal_stop(unsigned long flags)
> {
> struct sighand_struct *sighand = current->sighand;
>
> __set_current_state(TASK_STOPPED);
> spin_unlock_irqrestore(&sighand->siglock, flags);
> /* Don't run again until woken by SIGCONT or SIGKILL */
> freezable_schedule();
> }
Ah, understand. You think that we need to take ->siglock in advance
to avoid the race with SIGCONT?
No, we don't. Let me show you the code I suggested again:
void kthread_do_signal_stop(void)
{
spin_lock_irq(&curtent->sighand->siglock);
if (current->jobctl & JOBCTL_STOP_DEQUEUED)
__set_current_state(TASK_STOPPED);
spin_unlock_irq(¤t->sighand->siglock);
schedule();
}
so you can dequeue_signal() and call kthread_do_signal_stop() without
holding ->siglock. We can rely on JOBCTL_STOP_DEQUEUED bit. SIGCONT
clears it, so kthread_do_signal_stop() can't race.
Oleg.
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