Hi
Just do the binary calculus
0.0.254.255 is 0000 0000. 0000 0000. 1111 1110 . 1111 1111
apply that to your prefix 172.168.1.0
the first two bytes (172 and 168) will have to match
and the last bit of the 3rd byte needs to be 1
so that would be 172.168.1 .3 .5 .7 .8
in the other case .0 .2 .4 .6 you have a 0 at that position
These maches the routes to be denied (the odd ones)
And the even gets permitted by the permit any
In short
subnet masks need to be contiguous 1 ones followed by contiguous
zeros
wildcards don't 1010 0111 . 1111 0000 . 1110 0111 . 1010 1010 is a
valid wildcard ( 167.240.199.170)
regs
Haakon Claassen
EMEA - IT Transport Services -WAN
Cisco Systems
De Kleetlaan 6b - Pegasus Park
B-1831 Diegem (Belgium)
-----Original Message-----
From: Cisco Nuts [mailto:[EMAIL PROTECTED]]
Sent: zondag 29 september 2002 20:59
To: [EMAIL PROTECTED]
Subject: Filter odd routes via wildcard bits - How?? [7:54496]
Hello,
I am trying to understand how a wildcard mask of 0.0.254.255 filters odd
routes so that only even routes get across the router.
Ex. If you have routes for 172.168.1.0/24, 2.0/24, 3.0/24, 4.0/24,
5.0/24,
6.0/24 and you have an access-list of:
#access-list 11 deny 172.168.1.0 0.0.254.255
#access-list 11 permit any
And a:
#distribute-list 11 out
This will allow only: 172.168.2.0/24, 4.0/24 and 6.0/24 routes out.
Anyone can kindly explain the magic of this 0.0.254.255 wildcard?
Thank you.
Sincerely.
BTW: Anyone know the link on this on CCO?
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