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

[Hannes Frederic Sowa <hannes@stressinduktion.org>]

On 29.02.2016 16:19, Benjamin Poirier wrote:
> 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.

I think the code is fine albeit I think we should remove the min macro 
and just do something:

if (skb_tailroom(skb) > mtu)
	skb->reserved_tailroom = skb_tailroom(skb) - mtu;

Does that make sense? I think it is much more readable.

Thanks,
Hannes