On Thu, Oct 8, 2020 at 1:45 AM Xin Long <lucien....@gmail.com> wrote: > > On Thu, Oct 8, 2020 at 12:12 PM Cong Wang <xiyou.wangc...@gmail.com> wrote: > > > > skb_unshare() drops a reference count on the old skb unconditionally, > > so in the failure case, we end up freeing the skb twice here. > > And because the skb is allocated in fclone and cloned by caller > > tipc_msg_reassemble(), the consequence is actually freeing the > > original skb too, thus triggered the UAF by syzbot. > Do you mean: > frag = skb_clone(skb, GFP_ATOMIC); > frag = skb_unshare(frag) will free the 'skb' too?
Yes, more precisely, I mean: new = skb_clone(old) kfree_skb(new) kfree_skb(new) would free 'old' eventually when 'old' is a fast clone. The skb_clone() sets ->fclone_ref to 2 and returns the clone, whose skb->fclone is SKB_FCLONE_CLONE. So, the first call of kfree_skbmem() will just decrease ->fclone_ref by 1, but the second call will trigger kmem_cache_free() which frees _both_ skb's. Thanks.
