Depending upon your network (A, B, C, etc.) Your netmask would probably
be 255.255.255.0
255.255.248.0 ... etc. I don't think the last byte would ever be anything
but "0" unless you "own" a very small piece of the address. You want your
address "97" to fit into the mask, i.e., 0 to 97 inverted.
The kernel is now being "pickey" and actually checking these things. Try
255.255.255.0 even though you might not "own" 255 addresses. Just don't
use the ones you don't own.
It is now possible to get network allocations down to four addresses
-- i.e., a netmask of 255.255.255.252. It turns out that you can't
have an allocation of two addresses (255.255.255.254), because the
all-ones address on your network is reserved as a broadcast address,
and the all-zeroes address is reserved as the network number, so there
would be no actual host addresses in a two-address allocation.
Similarly, out of a four-address allocation, only two host addresses
are available, the ones ending in 01 and 10 binary.
Also, make sure that you are using the right netmask, not one "larger"
than your real allocation. If you attempt to send a packet to a host
that is within your netmask but not in your allocation, it won't go
through, because the kernel thinks it is accessible directly and won't
send it on to your gateway to the rest of the Internet.
Dale
-- Dale R. Worley Ariadne Internet Services Voice: +1 617-899-7949 Fax: +1 617-899-7946 E-mail: worley@ariadne.com "Internet-based electronic commerce solutions to real business problems."