On Sun, 23 Mar 2003 10:54:54 -0500 Kurt Wall <[EMAIL PROTECTED]> wrote:
> Hi, list,
>
> I've never been terribly clear on this, so I'll ask here. Given
> a network address of, say, 192.168.0.0 and a netmask of /8, thus
> 192.168.0.0/8, this means that 8 bits of the network address will
> be used for the host address, which means that any address in the
> range 192.168.0.1 - 192.168.0.255 will match. Am I correct?
>
you're backwards.
192.168.0.0/24 == 192.168.0.0-192.168.0.255
192.168.0.0/16 == 192.168.0.0-192.168.255.255
192.168.0.0/8 == 192.0.0.0-192.255.255.255
and
192.168.0.0/25 == 192.168.0.0-192.168.0.127
this is the VLSM subset of CIDR. The /# == the number of ones in the
netmask.
i.e., /8 == netmask 255.0.0.0, /24 == netmask 255.255.255.0, /25 =
netmask 255.255.255.128
(note: linewrap above at no additional charge)
Ciao,
David A. Bandel
--
Focus on the dream, not the competition.
Nemesis Racing Team motto
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