From: Scott Morris (swm@emanon.com)
Date: Sat Aug 13 2005 - 15:47:28 GMT-3
A '1' in the mask means that you don't care what the value is.
So looking at:
00000000 --> (starting value of the 3rd octet per 192.168.0.0)
00000001 --> mask you supplied
This only permits 0 or 1 in the 3rd octet.
You need a mask of 11111110 because the only one that MUST remain unchanged
is the 0 in the 1-bit position.
HTH,
Scott
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of Jody
Davis (joddavis)
Sent: Saturday, August 13, 2005 12:20 PM
To: Leigh Harrison; ccielab@groupstudy.com
Subject: RE: Even/Odd route filtering
As an example..
Filter only even routes from the 192.168.0.0..
Access-list 1 permit 192.168.0.0 0.0.1.0 (because the "3rd" octet of address
is "0" or an even, and the "1's" bit of the inverse mask is set to "don't
care", all evens will come through..
For odds,
Access-list 1 permit 192.168.1.0 0.0.1.0
Jody
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Leigh Harrison
Sent: Saturday, August 13, 2005 10:59 AM
To: ccielab@groupstudy.com
Subject: Even/Odd route filtering
All,
If I have the subnet 10.1.n.n - 10.50.n.n and I wanted to filter even routes
out of one interface and odd routes out of a different one, how would I
achieve this?
I know that all I would need to do is match on the 16th bit and ensure that
it is a "1" or a "0" in order to make it odd or even. I know that I could
write a big list, but I would like to match on the one bit. Is this even
possible ?!
Many thanks in advance,
LH
Leigh Harrison
Technical Consultant
Azzurri Communications
Castleton Mill,
Castleton Close,
Leeds,
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Email: Leigh.Harrison@azzu.co.uk <mailto:Leigh.Harrison@azzu.co.uk>
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