Re: [PATCH] sigqueue_free: fix the race with collect_signal()

[Sukadev Bhattiprolu <sukadev@us.ibm.com>]

Oleg Nesterov wrote:
> On 08/24, taoyue wrote:
>   
> > Oleg Nesterov wrote:
> >     
> > > > collect_signal:				sigqueue_free:
> > > >
> > > > 	list_del_init(&first->list);
> > > >                                       spin_lock_irqsave(lock, flags);
> > > >    
> > > >         
> > >                                         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> > >  
> > >       
> > > >                                       if (!list_empty(&q->list))
> > > >                                             list_del_init(&q->list);
> > > >                                       spin_unlock_irqrestore(lock, 
> > > >                                       flags);
> > > >                                       q->flags &= ~SIGQUEUE_PREALLOC;
> > > >
> > > >       __sigqueue_free(first);		__sigqueue_free(q);
> > > >    
> > > >         
> > > collect_signal() is always called under ->siglock which is also taken by
> > > sigqueue_free(), so this is not possible.
> > >
> > > Basically, this patch is the same one-liner I sent you before
> > >
> > > 	http://marc.info/?l=linux-kernel&m=118772206603453&w=2
> > >
> > > (Thanks for the additional testing and report, btw).
> > >
> > > P.S. It would be nice to know if this patch solves the problems reported
> > > by Jeremy, but his email is disabled.
> > >
> > > Oleg.
> > >
> > >  
> > >       
> > I know, using current->sighand->siglock to prevent one sigqueue
> > is free twice. I want to know whether it is possible that the two
> > function is called in different thread. If that, the spin_lock is useless.
> >     
> Not sure I understand. Yes, it is possible they are called by 2 different
> threads, that is why we had a race. But all threads in the same thread
> group have the same ->sighand, and thus the same ->sighand->siglock.
>   
Oleg, if one thread can be in collect_signal() and another in 
sigqueue_free() and both operate on the exact same sigqueue object, its 
not clear how we prevent two calls to __sigqueue_free() to
the same object. In that case the lock (or some lock) should be around 
__sigqueue_free() - no ?
i.e if we enter sigqueue_free(), we will call __sigqueue_free() 
regardless of the state.
> Oleg.
>
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