找到两个IPv4地址之间的网络距离（不是地理距离）

Question

Given an IPv4 address (needle) and an unsorted array of IPv4 addresses (haystack), how could I programmatically determine which single address in the given haystack is the closest (network-wise, NOT geographically) to the needle?

Since I don't have access to the netmask of every address, the solution should ignore netmasks and traceroute alike options.

All sorts of addresses are used, by which I mean: private, reserved, broadcast, lan and wan.

Any help in the form of theory, pseudo-code, python, php or perl is welcome.

The question Getting IP address list between two IP addresses is roughly similar, but it does quite cut it.

Answer 1

I'm still not quite sure what you're asking, but based on your comment

@PeterGibson by closest I meant, 192.168.1.101 is closer to 192.168.56.1 than to 172.30.130.66 . And 192.168.1.254 is closer to 192.168.1.240 than to 192.168.2.1

You could try the following python code for a distance function:

import socket
def dist(a, b):
    def to_num(addr):
        # parse the address string into integer quads
        quads = map(ord, socket.inet_aton(addr))
        # spread the quads out 
        return reduce(lambda x,y: x * 0x10000 + y, quads)
    return abs(to_num(a) - to_num(b))

The numbers returned are fairly arbitrary, but should suffice for basic needs. I'm still not sure what it should do with broadcast addresses etc though.

Some examples:

>>> dist('192.168.1.254', '192.168.1.240')
14L
>>> dist('192.168.1.254', '192.168.2.1')
65283L
>>> dist('192.168.1.101', '192.168.56.1')
3604380L
>>> dist('192.168.1.101', '172.30.130.66')
5630092231245859L