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

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

Oleg Nesterov wrote:
> On 08/24, Sukadev Bhattiprolu wrote:
>   
> > 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.
> > > > >
> > > > >
> > > > >      
> > > > >           
> > > > 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.
> >     
>
> Yes. They both will call __sigqueue_free(). But please note that __sigqueue_free()
> checks SIGQUEUE_PREALLOC, which is cleared by sigqueue_free().
>
> IOW, when sigqueue_free() unlocks ->siglock, we know that it can't be used
> by collect_signal() from another thread. So we can clear SIGQUEUE_PREALLOC
> and free sigqueue. We don't need this lock around sigqueue_free() to prevent
> the race. collect_signal() can "see" only those sigqueues which are on list.
>
> IOW, when sigqueue_free() takes ->siglock, colect_signal() can't run, because
> it needs the same lock. Now we delete this sigqueue from list, nobody can
> see it, it can't have other references. So we can unlock ->siglock, mark
> sigqueue as freeable (clear SIGQUEUE_PREALLOC), and free it.
>
> Do you agree?
>   

Yes. I see it now. I had missed the SIGQUEUE_PREALLOC in __sigqueue_free().

Thanks for clarifying

Suka


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