On 08/24, Sukadev Bhattiprolu wrote:
quoted text
>
> 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?

Oleg.

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