Python：在IP地址八位字节之间执行数学功能

Question

I'm trying to perform the following:

firstoctet * 256 + secondoctet = * 256 + thirdoctet = * 256 + fourthoctet = x

When I use this as an example :

64.233.187.99 (google.com)
64 * 256 + 233 = * 256 + 187 = * 256 + 99 = http://1089059683/

Can someone please provide a method as to how this can be done? The mathematical sequence is no problem, i'm just unsure as to how I can take the octet values out of the decimal points to perform the math functions.

Thanks in advance!

Answer 1

If you're on Python 3.3 or higher, you can leverage the ipaddress module to avoid reinventing the wheel, and it even provides a useful view as bytes that int.from_bytes can efficiently convert to a real int :

from ipaddress import IPv4Address

ip_as_addr = IPv4Address("64.233.187.99")
ip_as_int = int.from_bytes(ip_as_addr.packed, 'big')
print(ip_as_int, hex(ip_as_int))

gets you output of 1089059683 0x40e9bb63 .

You could do this by hand if you really wanted to, I just like the self-documenting aspect of the above code. If Py 3.3+ isn't an option, you can get the same results with:

octets = map(int, "64.233.187.99".split('.'))
ip_as_int = sum(octet << ((3 - i) * 8) for i, octet in enumerate(octets))

That just splits the octets apart, converts them to int , then shifts each of them left by 24, 16, 8 and 0 bits (to align the octets properly), which then allows sum to combine them into a single int .

Answer 2

Convert an IP address string to a list of integers like this:

ip_as_string = "64.233.187.99"
ip_as_ints = [int(a) for a in ip_as_string.split('.')]

ip_as_ints will be [64, 233, 187, 99].

I will guess that the mathematical expression that you intend to perform is RPN for converting the byte sequence from a base-256 representation to a single number. (Thanks to ShadowRanger for showing this). You can obtain that single number easily as follows:

x = functools.reduce(lambda x,y: (x << 8) | y, ip_as_ints, 0)
print(x)

x is 1089059683.