Python Network/cidr Calculations

Question

I am working on building an embedded network appliance (linux based) and have come across the need to dynamically build daemon conf files. As such, I need to be able to do some network address calculations in the python code that will be building the conf files. I am not a programmer so I'm fearful I wrote a module that will fail to function as I'd hoped once the appliance starts shipping.

Below is what I have so far, its really pieced together with what I could find on this site and Google.

Is there a better method to find the network address and cidr for a network interface? Converting the netmask to a bin str and counting the 1's seems pretty inelegant.

import socket
import fcntl
import struct

SIOCGIFNETMASK = 0x891b
SIOCGIFADDR = 0x8915

s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)

def _GetIfaceMask(iface):
    return struct.unpack('L', fcntl.ioctl(s, SIOCGIFNETMASK, struct.pack('256s', iface))[20:24])[0]

def _GetIfaceAddr(iface):
    return struct.unpack('L', fcntl.ioctl(s, SIOCGIFADDR, struct.pack('256s', iface[:15]))[20:24])[0]

def GetIfaceNet(iface):
    net_addr = _GetIfaceAddr(iface) & _GetIfaceMask(iface)
    return socket.inet_ntoa(struct.pack('L', net_addr))

def GetIfaceCidr(iface):
    bin_str = bin(_GetIfaceMask(iface))[2:]
    cidr = 0
    for c in bin_str:
        if c == '1':  cidr += 1
    return cidr

Thanks for any input, I really am somewhat lost on this. If this isn't the place for this type of feedback, please let me know.

Answer 1

You may check iptools python module http://code.google.com/p/python-iptools/ it can convert from long to dotted ip format and vise versa.

Answer 2

This can be solved using a Hamming weight algorithm. Stolen from How to count the number of set bits in a 32-bit integer? and translated into Python:

def number_of_set_bits(x):
    x -= (x >> 1) & 0x55555555
    x = ((x >> 2) & 0x33333333) + (x & 0x33333333)
    x = ((x >> 4) + x) & 0x0f0f0f0f
    x += x >> 8
    x += x >> 16
    return x & 0x0000003f

Another, more readable solution (but running in O(log x)):

def number_of_set_bits(x):
    n = 0
    while x:
        n += x & 1
        x = x >> 1
    return n