On 2016/02/29 15:57, Daniel Borkmann wrote: [...] > > [ cutting the IPv4 part off as diff is the same ] > > >diff --git a/net/ipv6/mcast.c b/net/ipv6/mcast.c > >index 5ee56d0..c157edc 100644 > >--- a/net/ipv6/mcast.c > >+++ b/net/ipv6/mcast.c > >@@ -1574,9 +1574,9 @@ static struct sk_buff *mld_newpack(struct inet6_dev > >*idev, unsigned int mtu) > > return NULL; > > > > skb->priority = TC_PRIO_CONTROL; > >- skb->reserved_tailroom = skb_end_offset(skb) - > >- min(mtu, skb_end_offset(skb)); > > skb_reserve(skb, hlen); > >+ skb->reserved_tailroom = skb_tailroom(skb) - > >+ min_t(int, mtu, skb_tailroom(skb) - tlen); > > Are you sure this is correct? Wouldn't that mean (assuming we allocated > enough space), that I could now fill a larger than MTU frame?
Quoting back a part of the log: > >The maximum space available for ip headers and payload without > >fragmentation is min(mtu, data + extra). Therefore, > >reserved_tailroom > >= data + extra + tlen - min(mtu, data + extra) > >= skb_end_offset - hlen - min(mtu, skb_end_offset - hlen - tlen) > >= skb_tailroom - min(mtu, skb_tailroom - tlen) ; after skb_reserve(hlen) The min() takes care of the situation you describe, ie. if the allocated space is large, reserved_tailroom will be large enough that we do not use more space than the mtu. I tested the mld and igmp code with different driver parameters, mtu values, number of multicast address records and even allocation failures. If you think the formula is wrong, please provide a counter-example with hlen, tlen, mtu and size values. Regards, -Benjamin
