> However, for 192.168.0.0 with a subnet mask of 255.255.254.0, then
> 192.168.1.255 is the broadcast address, with 192.168.1.254 being the last
> useable address and 192.168.0.1 being the first. In this example,
> 192.168.0.255 is a perfectly legal IP address.
>
> Of course, I could be completely wrong. Somebody hit me if I am =]
>

you're absolutely correct - the broadcast address is entirely dependent
on the subnet mask. Any netmask of 23 bits or less will result in an
address of x.x.x.255 is a valid address (assuming there's at least one 0
bit going left from bit 23..)

I'll hit you anyway if it makes you feel better though :)

--
Ian Batterbee
Senior Network/Comms Technician (IT Group)
Auckland University of Technology
Auckland, New Zealand
Mobile: +64-21-655-625
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