When in doubt, do it in binary:
0 = 00000000
64 = 01000000
128 = 10000000
192 = 11000000
As a /26 (taking as given the 24 x's not shown for the other three octets):
xxyyyyyy - y's are the host
xx doesn't have the same value from 64-191
As a /25:
xyyyyyy - y's are the host
x doesn't have the same value from 64-191
A /26 means there are 26 bits of network and 6 bits of host; a /25 means 25
bits of network and 7 for host.
As /26's xx = {00, 01, 10, 11} for the four distinct values, with the other
six bits signifying the host address. As /25s x = {0, 1} are the two
distinct values. Now, you could use 0/25 (as 0-127 all have same first bit)
and 128/26 and 192/26 as each of these network address bits (10 and 11) stay
the same through the respective block of addresses.
Hopefully this helps.
Clark
> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Christoffer Pio
> Sent: Wednesday, August 27, 2003 10:04 AM
> To: [EMAIL PROTECTED]
> Subject: subnetting C class into /26 /25 /26, why can this be done?
>
>
> Is it not possible to subnet a C class into 3 nets, like
>
> 0-63
> 64-191 <-- Offending network (?)
> 192-255
>
> If so, why is this?
>
> Christoffer
> _______________________________________________
> [EMAIL PROTECTED] mailing list
> http://lists.freebsd.org/mailman/listinfo/f> reebsd-net
> To
> unsubscribe, send any mail to
> "[EMAIL PROTECTED]"
>
_______________________________________________
[EMAIL PROTECTED] mailing list
http://lists.freebsd.org/mailman/listinfo/freebsd-net
To unsubscribe, send any mail to "[EMAIL PROTECTED]"
