begin quoting markw as of Fri, Jun 06, 2008 at 03:57:03PM -0700: > Karl Cunningham wrote: > >$ ipcalc 167.190.1.0 255.255.248.0 > >Address: 167.190.1.0 10100111.10111110.00000 001.00000000 > >Netmask: 255.255.248.0 = 21 11111111.11111111.11111 000.00000000 > >Wildcard: 0.0.7.255 00000000.00000000.00000 111.11111111 > >=> > >Network: 167.190.0.0/21 10100111.10111110.00000 000.00000000 > >HostMin: 167.190.0.1 10100111.10111110.00000 000.00000001 > >HostMax: 167.190.7.254 10100111.10111110.00000 111.11111110 > >Broadcast: 167.190.7.255 10100111.10111110.00000 111.11111111 > >Hosts/Net: 2046 Class B > > You can do this all in your head. Just subtract the mask from 256. > 256-248 is 8, so you have 0-7, with 0.0 as the network, 7.255 as the > broadcast and a whole lot of hosts and noise in between.
Careful. The netmask doesn't HAVE to be contiguous. > The network portion is extracted by logically anding the netmask with > the address. This example makes it easy to see that. Again, just > remember 256 and you can usually figure it out before you can punch it > into ipcalc, or some similar tool. Do it a few times manually (by converting to binary and back), then use a tool, and most of the time you don't need to worry about such things except when the network is acting up. -- I use bc with obase=2 and ibase=10 (or vice versa). Stewart Stremler -- KPLUG-List@kernel-panic.org http://www.kernel-panic.org/cgi-bin/mailman/listinfo/kplug-list
