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Re: [Openvpn-users] Duplicate MAC addresses with OSX clients


  • Subject: Re: [Openvpn-users] Duplicate MAC addresses with OSX clients
  • From: Andrew Boyce-Lewis <aboyce@xxxxxxxxxxxxxx>
  • Date: Wed, 5 Oct 2005 16:03:27 -0400

I'd actually read that a few days ago in the source :-)

I'm all for adding some similar functionality to the tun/tap OSX driver (with appropriate warning), I just am having trouble figuring out exactly where to do it (I'm a very very rusty c++ programmer). 


Andrew E. Boyce-Lewis

System and Network Administrator

Conduit Internet Technologies, Inc.

800.493.5045 x 210

814.867.8248 Fax

http://www.conduit-it.com

aboyce@xxxxxxxxxxxxxx




On Oct 5, 2005, at 3:56 PM, James Yonan wrote:



On Wed, 5 Oct 2005, Andrew Boyce-Lewis wrote:


I am running a bridging network and have discovered that the OSX tun/ 
tap driver apparently hard codes the MAC address of the tap interface  
(so all OSX clients have the same address). ( Tiger version http:// 
www-user.rhrk.uni-kl.de/~nissler/tuntap/ )

Has anyone else run into this problem? or can anyone suggest any  
possible workarounds? I've read that the windows and linux tap  
drivers use an algorithm to generate a non-sense MAC address and  
assign that to the interface... I suppose that might need to be  
implemented for the OSX version.


Here are my comments in the TAP-Win32 source code about this.

/*
 * Generate a MAC using the GUID in the adapter name.
 *
 * The mac is constructed as 00:FF:xx:xx:xx:xx where
 * the Xs are taken from the first 32 bits of the GUID in the
 * adapter name.  This is similar to the Linux 2.4 tap MAC
 * generator, except linux uses 32 random bits for the Xs.
 *
 * In general, this solution is reasonable for most
 * applications except for very large bridged TAP networks,
 * where the probability of address collisions becomes more
 * than infintesimal.
 *
 * Using the well-known "birthday paradox", on a 1000 node
 * network the probability of collision would be
 * 0.000116292153.  On a 10,000 node network, the probability
 * of collision would be 0.01157288998621678766.
 */

James


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