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Message-ID: <20160229151920.GA5474@f1.synalogic.ca>
Date: Mon, 29 Feb 2016 07:19:20 -0800
From: Benjamin Poirier <bpoirier@...e.com>
To: Daniel Borkmann <daniel@...earbox.net>
Cc: netdev@...r.kernel.org, Eric Dumazet <eric.dumazet@...il.com>,
Hannes Frederic Sowa <hannes@...essinduktion.org>,
Hideaki YOSHIFUJI <yoshfuji@...ux-ipv6.org>
Subject: Re: [PATCH] mld, igmp: Fix reserved tailroom calculation
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
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