Messages in this thread | | | Date | Sat, 25 Aug 2007 10:24:44 -0700 | From | Sukadev Bhattiprolu <> | Subject | Re: [PATCH] sigqueue_free: fix the race with collect_signal() |
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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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