Date: 18 Dec 2001 16:09:28 -0800 From: Joel Ray Holveck <joelh@gnu.org> To: freebsd-questions@freebsd.org Subject: IPSec: Is it possible to require tunnel AH? Message-ID: <87y9k0vznr.fsf@thor.piqnet.org>
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I've been having trouble with the use of AH in a tunnel context. I
don't see any way to require AH on incoming packets with IPv4.
Consider the following fragment:
spdadd 192.168.13.8/32 0.0.0.0/0 any -P in ipsec
ah/tunnel/192.168.13.8-192.168.13.2/use ;
If I switch 'use' to 'require', then all packets are rejected. I can
give you more context if you like, but I didn't want to make my
question overly long; in a nutshell, the 192.168.13.8 box is on an
802.11 network.
Note that the handbook uses this in an example:
http://www.freebsd.org/doc/en_US.ISO8859-1/books/handbook/ipsec.html
in the section on IPv4 tunnel mode.
If there is no 'require' ability, then what is the use of tunnel mode
AH? In this instance, I don't care if the rest of the Internet had
done icky stuff before the tunnel endpoint, because I can filter that
out at or before said endpoint.
I think this hinges around the following bit of code, from
/sys/netinet6/ah_input.c:
#if 1
/*
* Should the inner packet be considered authentic?
* My current answer is: NO.
*
* host1 -- gw1 === gw2 -- host2
* In this case, gw2 can trust the authenticity of the
* outer packet, but NOT inner. Packet may be altered
* between host1 and gw1.
*
* host1 -- gw1 === host2
* This case falls into the same scenario as above.
*
* host1 === host2
* This case is the only case when we may be able to leave
* M_AUTHIPHDR and M_AUTHIPDGM set.
* However, if host1 is wrongly configured, and allows
* attacker to inject some packet with src=host1 and
* dst=host2, you are in risk.
*/
m->m_flags &= ~M_AUTHIPHDR;
m->m_flags &= ~M_AUTHIPDGM;
#endif
Thanks,
joelh
--
Joel Ray Holveck - joelh@gnu.org
Fourth law of programming:
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