Network Address/Netmask Notation

[Kurt Wall]

An unnamed Administration source, David A. Bandel, wrote:
% On Sun, 23 Mar 2003 10:54:54 -0500
% Kurt Wall <kwall at kurtwerks.com> 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

Thanks, David.

% (note: linewrap above at no additional charge)

Feature!

Kurt
-- 
"He was a modest, good-humored boy.  It was Oxford that made him
insufferable."