An unnamed Administration source, David A. Bandel, wrote: % 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
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." _______________________________________________ Linux-users mailing list [EMAIL PROTECTED] Unsubscribe/Suspend/Etc -> http://www.linux-sxs.org/mailman/listinfo/linux-users
