Problem with MIT Kerberos v1.4, OpenSSH 3.9p1 and Active Directory
Jeffrey Altman
jaltman2 at nyc.rr.com
Thu Feb 10 13:39:58 EST 2005
You can perform a network trace from AD with netmon.exe to see
whether or not you are using TCP. You should be otherwise you
would not get a response. If you are getting ASN.1 ending
unexpectedly it sounds like a buffer is being truncated somewhere.
Jeffrey Altman
Sam Evans wrote:
> All:
>
> I seem to have run into a road block getting my Linux machines to
> authenticate against AD when coming in through OpenSSH.
>
> First, let me start off my listing what my environmnet is:
>
> Test Client:
> * RHEL Linux
> * MIT Kerboros v1.4
> * OpenSSH v3.9p1 - Compiled using the following line:
> ../configure --with-tcp-wrappers --with-pam
> --with-kerberos5=/usr/kerberos --with-md5-passwords --prefix=/usr
> --sysconfdir=/etc/ssh
>
> Active Directory:
> * Windows 2003
>
> Scenario 1:
>
> If I use my local account and password, I can get into the machine OK. I
> know that OpenSSH is functioning properly. At this point, if I do a
> 'kinit' I can successfully authenticate myself against AD and obtain my
> Keberos5 ticket.
>
> Scenario 2:
>
> If I change my account information to require that authentication take
> place using Kerberos, then I get the following error from the ssh daemon:
>
> debug1: Kerberos password authentication failed: ASN.1 encoding ended
> unexpectedly
>
> -- What I have been able to determine at this point is that if I remove
> my userid from the multitude of groups that it belongs to in AD, then I
> *can* successfully authenticate myself when I come in through OpenSSH,
> using Kerberos.
>
> -- If I place myself back into the same groups, I cannot authenticate
> myself and get the above error.
>
> In doing some reading, it appears as if I need to force TCP usage in the
> MIT Kerberos, which I have done. Everything still works when I do
> 'kinit' but nothing has changed in regards to OpenSSH authentication
> ability.
>
> Anyone have any thoughts or suggestions?
>
> Thanks,
> Sam
> P
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