Get the IP address of a network interface in Python 3

get_ip_address.py

#!/usr/bin/env python3
# https://gist.github.com/EONRaider/3b7a8ca433538dc52b09099c0ea92745
__author__ = 'EONRaider, keybase.io/eonraider'

import fcntl
import socket
import struct

try:
    from netifaces import AF_INET, ifaddresses
except ModuleNotFoundError as e:
    raise SystemExit(f"Requires {e.name} module. Run 'pip install {e.name}' "
                     f"and try again.")


def get_ip_linux(interface: str) -> str:
    """
    Uses the Linux SIOCGIFADDR ioctl to find the IP address associated
    with a network interface, given the name of that interface, e.g.
    "eth0". Only works on GNU/Linux distributions.
    Source: https://bit.ly/3dROGBN

    Returns:
        The IP address in quad-dotted notation of four decimal integers.
    """

    sock = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)
    packed_iface = struct.pack('256s', interface.encode('utf_8'))
    packed_addr = fcntl.ioctl(sock.fileno(), 0x8915, packed_iface)[20:24]
    return socket.inet_ntoa(packed_addr)


def get_ip_cross(interface: str) -> str:
    """
    Cross-platform solution that should work under Linux, macOS and
    Windows.    
    """

    return ifaddresses(interface)[AF_INET][0]['addr']

Thanks!

For get_ip_linux(), maybe with statement is even better, as it would close the socket upon completion or exception:

with socket.socket(socket.AF_INET, socket.SOCK_DGRAM) as ssock: