From: Darek Kuzma (darekk@xxxxxxxxxxxxx)
Date: Wed Jul 17 2002 - 14:47:52 GMT-3
>>> Assuming you have to permit everything else, try .... <<<
is incorrect!!!
140.199.57.0 45.0.0.0 will match more than networks required !!!
for example:
129.199.57.0
136.199.57.0
141.199.57.0
160.199.57.0
etc...
Darek
John Mistichelli wrote:
> Sure,
>
> For the first two addresses only the first octet
> needed wildcard bits. Convert to binary:
> 140 = 10001100
> 161 = 10100001
>
> Assuming a number line of 128,64,32,16,8,4,2,1 the
> bits that are not in common between the two are 32, 8,
> 4 and 1. If you just add those together you get 45.
> All other octets had to match exactly. Hence 45.0.0.0.
>
> I used the same method on the other two only setting
> wilcard bits in the second octet.
>
> 59 = 00111011
> 63 = 00111111
>
> All bits are common except the 4 position. Hence
> 0.4.0.0.
>
> Hope that helps.
> John
> 7536
This archive was generated by hypermail 2.1.4 : Sat Sep 07 2002 - 19:36:34 GMT-3