convert cidr to subnet mask in python

Question

I have a python dictionary that I've created, this dictionary contains a list of subnets in the following format:

x.x.x.x/24
y.y.y,y/25
z.z.z.z/26
a.a.a.a/27 

etc...

I would like to take the items in this dictionary, parse it, then spit out the results in the following format:

x.x.x.x 255.255.255.0
y.y.y.y 255.255.255.128
x.x.x.x 255.255.255.192
a.a.a.a 255.255.255.224 

I don't have much on this as of right now because I can't find a lot on this topic on the web, not anything that can be in a quick and concise way that is. Thoughts?

Angulo

Your data is not dictionary as you mentioned! – Anglice

If it helps to see an example, I made a class that does this here. – Nodababus

Answer 1

Code:

import socket
import struct

def cidr_to_netmask(cidr):
    network, net_bits = cidr.split('/')
    host_bits = 32 - int(net_bits)
    netmask = socket.inet_ntoa(struct.pack('!I', (1 << 32) - (1 << host_bits)))
    return network, netmask

Usage:

>>> cidr_to_netmask('10.10.1.32/27')
('10.10.1.32', '255.255.255.224')
>>> cidr_to_netmask('208.128.0.0/11')
('208.128.0.0', '255.224.0.0')
>>> cidr_to_netmask('208.130.28.0/22')
('208.130.28.0', '255.255.252.0')

Paphos

Goes without saying, but for the IPv6 initiated. CIDR is the official notation for IPv6, there for netmask is not a thing anymore. But if you still wanted to, you could do it: gist.github.com/Torxed/148303f1677e56dcca794386e90299fc – Postcard

Answer 2

I thought I'd throw in my own solution, since I was going for a little bit more readability in it than the other answers shown.

def cidr_to_netmask(cidr):
  cidr = int(cidr)
  mask = (0xffffffff >> (32 - cidr)) << (32 - cidr)
  return (str( (0xff000000 & mask) >> 24)   + '.' +
          str( (0x00ff0000 & mask) >> 16)   + '.' +
          str( (0x0000ff00 & mask) >> 8)    + '.' +
          str( (0x000000ff & mask)))

It's now easier to see what's going on, and that is:

1. Get the numeric mask by padding the front with 1s and having the cidr make up the rest
2. For each bit, apply the mask
3. Concatenate all the bits together, separated by periods

This is very procedural and does not use any libraries.

Elvaelvah

Answer 3

Try this solution:

Python3

from ipaddress import IPv4Network

networks = {'n1':'10.1.0.0/21','n2':'10.2.0.0/22','n3':'10.3.0.0/23','n4':'10.4.0.0/24'}

for x,y in enumerate(networks):
    print(IPv4Network(networks[y]).network_address, IPv4Network(networks[y]).netmask)

Result:

10.1.0.0 255.255.248.0
10.2.0.0 255.255.252.0
10.3.0.0 255.255.254.0
10.4.0.0 255.255.255.0

Python2

from netaddr import IPNetwork

networks = {'n1':'10.1.0.0/21','n2':'10.2.0.0/22','n3':'10.3.0.0/23','n4':'10.4.0.0/24'}

for x,y in enumerate(networks):
    print(str(IPNetwork(networks[y]).network), str(IPNetwork(networks[y]).netmask))

Result:

('10.1.0.0', '255.255.248.0')
('10.2.0.0', '255.255.252.0')
('10.3.0.0', '255.255.254.0')
('10.4.0.0', '255.255.255.0')

Purnell

apt install python-netaddr – Gustafson

pypi.org/project/ipaddress is a "Port of the 3.3+ ipaddress module to 2.6, 2.7, 3.2" in case you still need to maintain Python2 code and want to use ipaddress. – Triptych

Answer 4

Try this

lsIP = []
ans = 0
CIDR = 32
IP = [1] * CIDR
for i in range(len(IP)):
    iIdx = i % 8
    if iIdx == 0:
        if i >= 8:
            lsIP.append(ans)
            ans = 0
    ans += pow(2, 7 - iIdx)
lsIP.append(ans)

[lsIP.append(0) for i in range(4 - len(lsIP))]
print lsIP

Cantor

Answer 5

Here is one I made:

def cidr_to_netmask(cidr) -> str:
    cidr = int(cidr)
    mask_bin = f"{'1'*cidr}{'0'*(32-cidr)}"
    out = f"{int(mask_bin[0:8], 2)}.{int(mask_bin[8:16], 2)}.{int(mask_bin[16:24], 2)}.{int(mask_bin[24:32], 2)}"
    return out

Masera