pf: Fix possible incorrect IPv6 fragmentation

When forwarding pf tracks the size of the largest fragment in a fragmented
packet, and refragments based on this size.
It failed to ensure that this size was a multiple of 8 (as is required for all
but the last fragment), so it could end up generating incorrect fragments.

For example, if we received an 8 byte and 12 byte fragment pf would emit a first
fragment with 12 bytes of payload and the final fragment would claim to be at
offset 8 (not 12).

We now assert that the fragment size is a multiple of 8 in ip6_fragment(), so
other users won't make the same mistake.

Reported by:	Antonios Atlasis <aatlasis at secfu net>
MFC after:	3 days

(cherry picked from commit 4f3397263b95a45dd58e2be3a566029f8841cace)

2 files changed, 6 insertions, 0 deletions

diff --git a/sys/netinet6/ip6_output.c b/sys/netinet6/ip6_output.c
index 09eab20..7af04b0 100644
--- a/sys/netinet6/ip6_output.c
+++ b/sys/netinet6/ip6_output.c
@@ -219,6 +219,8 @@ ip6_fragment(struct ifnet *ifp, struct mbuf *m0, int hlen, u_char nextproto,
 	int error;
 	int tlen = m0->m_pkthdr.len;
 
+	KASSERT(( mtu % 8 == 0), ("Fragment length must be a multiple of 8"));
+
 	m = m0;
 	ip6 = mtod(m, struct ip6_hdr *);
 	mnext = &m->m_nextpkt;
diff --git a/sys/netpfil/pf/pf_norm.c b/sys/netpfil/pf/pf_norm.c
index 3559909..ead119f 100644
--- a/sys/netpfil/pf/pf_norm.c
+++ b/sys/netpfil/pf/pf_norm.c
@@ -1128,6 +1128,10 @@ pf_refragment6(struct ifnet *ifp, struct mbuf **m0, struct m_tag *mtag)
 		hdr->ip6_nxt = IPPROTO_FRAGMENT;
 	}
 
+	/* The MTU must be a multiple of 8 bytes, or we risk doing the
+	 * fragmentation wrong. */
+	maxlen = maxlen & ~7;
+
 	/*
 	 * Maxlen may be less than 8 if there was only a single
 	 * fragment.  As it was fragmented before, add a fragment