Messages in this thread | | | Date | Mon, 15 Jun 2015 21:14:29 +0200 | From | Oleg Nesterov <> | Subject | Re: [RFC PATCH 06/18] signal/kthread: Initial implementation of kthread signal handling |
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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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