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

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On 08/23, Sukadev Bhattiprolu wrote:
>
> Oleg Nesterov wrote:
> >Spotted by taoyue <[email protected]> and Jeremy Katz 
> ><[email protected]>.
> >
> >collect_signal:				sigqueue_free:
> >
> >	list_del_init(&first->list);
> >						if (!list_empty(&q->list)) {
> >							// not taken
> >						}
> >						q->flags &= 
> >						~SIGQUEUE_PREALLOC;
> >
> >	__sigqueue_free(first);			__sigqueue_free(q);
> >
> >Now, __sigqueue_free() is called twice on the same "struct sigqueue" with 
> >the
> >obviously bad implications.
> >
> >--- t/kernel/signal.c~SQFREE	2007-08-22 20:06:31.000000000 +0400
> >+++ t/kernel/signal.c	2007-08-23 16:02:57.000000000 +0400
> >@@ -1297,20 +1297,19 @@ struct sigqueue *sigqueue_alloc(void)
> > void sigqueue_free(struct sigqueue *q)
> > {
> > 	unsigned long flags;
> >+	spinlock_t *lock = &current->sighand->siglock;
> >+
> > 	BUG_ON(!(q->flags & SIGQUEUE_PREALLOC));
> > 	/*
> > 	 * If the signal is still pending remove it from the
> >-	 * pending queue.
> >+	 * pending queue. We must hold ->siglock while testing
> >+	 * q->list to serialize with collect_signal().
> > 	 */
> >-	if (unlikely(!list_empty(&q->list))) {
> >-		spinlock_t *lock = &current->sighand->siglock;
> >-		read_lock(&tasklist_lock);
> >-		spin_lock_irqsave(lock, flags);
> >  
> Hmm, but the existing code _does_ take the siglock here. Is that not 
> sufficient ?

Yes, it does, and this is sufficient, so the patch removes tasklist_lock.

> Isn't the first list_empty() check without lock only an optimization for 
> the common
> case ?

Yes, this is optimization, but I strongly believe it is wrong. Please look
at the race description above.

!list_empty(&q->list) does _not_ necessary mean that q is not used and we can
free it. It is possible that collect_signal() just removed this sigqueue from
list (list_empty(&q->list) becomes true) and going to free it.

The whole point is: we can't check list_empty() without ->siglock, this is
racy, and leads to double-free.

Oleg.

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