It's likely that the subnet masks are throwing you off. As long as you keep in mind that the below rules no longer apply, you should be fine.
Ultimately classful addressing came down to the most significant (or "leading") bits in the address. Nothing more, nothing less.
- Class A: Most significant bits starts with
0
- Class B: Most significant bits start with
10
- Class C: Most significant bits start with
110
The "classes" came from the way they split up the address space for use between "host" and "network". Keep in mind that back then (way way back, from the days of ARPANET), subnet masks did not exist, and the network was intended to be inferred from the address itself. So, with the above in mind, this is what they came up with (this is intended to be binary representation - each N or H represents a single bit in the 32-bit address):
- Class A:
NNNNNNNN.HHHHHHHH.HHHHHHHH.HHHHHHHH (less networks, more hosts)
- Class B:
NNNNNNNN.NNNNNNNN.HHHHHHHH.HHHHHHHH (more networks, less hosts)
- Class C:
NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH (even more networks, even less hosts)
Here the N is representative of the network portion of the address, and the H is representative of the host portion of the address, or as they called it back in the day, the "rest field."
Combining that with what was said earlier about the most-significant bits, we have the following:
- Class A: 0.0.0.0 - 127.255.255.255
- Class B: 128.0.0.0 - 191.255.255.255
- Class C: 192.0.0.0 - 223.255.255.255
Converting those ranges to binary may make this more clear:
Class A
0.0.0.0
-----------
[0]0000000.00000000.00000000.00000000
127.255.255.255
-----------
[0]1111111.11111111.11111111.11111111
^
most significant bit = 0
Class B
128.0.0.0
-----------
[10]000000.00000000.00000000.00000000
191.255.255.255
-----------
[10]111111.11111111.11111111.11111111
^
most significant bits = 10
Class C
192.0.0.0
-----------
[110]00000.00000000.00000000.00000000
223.255.255.255
-----------
[110]11111.11111111.11111111.11111111
^
most significant bits = 110
Every single address within those ranges will share a common leading bit(s). The moral of the story is, if you can remember what the leading bits are supposed to be (0 for class A, 10 for class B, 110 for class C) it's extremely simple to determine what "class" an address would have otherwise belonged in. Or, if decimal is easier:
- Class A: First octet in address is between 0 and 127, inclusive
- Class B: First octet in address is between 128 and 191, inclusive
- Class C: First octet in address is between 192 and 223, inclusive
The easiest way to mess someone up on "classful addressing" either on a test, or exam, or whatever, is to use misdirection by way of a subnet mask. Again, remember that the subnet mask does not apply for determining the class of an address. This is easy to forget because as others have said, classless addressing and routing have been around for over two decades now, and the subnet mask and CIDR notation have become ubiquitous in the industry.
It doesn't answer your question, but may answer your need:
I fear the only option would be to have all peers register their local IP addresses on a server directly connected to the Internet (without NAT).
Each peer could then query the server for the local and translated IP addresses of others. If their translated public IP addresses are in the same range (allocated in the same AS), they could then try a direct connection through their local IP addresses.
Of course, with end-to-end IPv6 connectivity, there would be no issue...
Best Answer
IPv6 doesn't have NAT. NAT is a hack or kludge in order to forestall the depletion of IPv4 addresses until IPv6 could become ubiquitous. NAT breaks the end-to-end connectivity which is fundamental to the IP design. IPv6 restores the end-to-end connectivity.