Re: [PATCH] mld, igmp: Fix reserved tailroom calculation

[Benjamin Poirier <bpoirier@suse.com>]

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