appletalk: Fix Use-After-Free in atalk_ioctl
Because atalk_ioctl() accesses sk->sk_receive_queue
without holding a sk->sk_receive_queue.lock, it can
cause a race with atalk_recvmsg().
A use-after-free for skb occurs with the following flow.
```
atalk_ioctl() -> skb_peek()
atalk_recvmsg() -> skb_recv_datagram() -> skb_free_datagram()
```
Add sk->sk_receive_queue.lock to atalk_ioctl() to fix this issue.
Fixes: 1da177e4c3
("Linux-2.6.12-rc2")
Signed-off-by: Hyunwoo Kim <v4bel@theori.io>
Link: https://lore.kernel.org/r/20231213041056.GA519680@v4bel-B760M-AORUS-ELITE-AX
Signed-off-by: Paolo Abeni <pabeni@redhat.com>
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@ -1775,15 +1775,14 @@ static int atalk_ioctl(struct socket *sock, unsigned int cmd, unsigned long arg)
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break;
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}
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case TIOCINQ: {
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/*
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* These two are safe on a single CPU system as only
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* user tasks fiddle here
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*/
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struct sk_buff *skb = skb_peek(&sk->sk_receive_queue);
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struct sk_buff *skb;
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long amount = 0;
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spin_lock_irq(&sk->sk_receive_queue.lock);
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skb = skb_peek(&sk->sk_receive_queue);
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if (skb)
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amount = skb->len - sizeof(struct ddpehdr);
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spin_unlock_irq(&sk->sk_receive_queue.lock);
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rc = put_user(amount, (int __user *)argp);
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break;
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}
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