Getting a random free tcp port in python using sockets

get-free-tcp-port.py

# Getting a random free tcp port in python using sockets

def get_free_tcp_port():
    tcp = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    tcp.bind(('', 0))
    addr, port = tcp.getsockname()
    tcp.close()
    return port


def get_free_tcp_address():
    tcp = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    tcp.bind(('', 0))
    host, port = tcp.getsockname()
    tcp.close()
    return 'tcp://{host}:{port}'.format(**locals())

This is useful for showing that

• 0 as the socket number lets the OS select the port
• getsocketname() will return the address.

But beyond that you should never do it this way. As @corydodt points out there's a race condition. Rather like creating a temporary file you should both allocate the address (including port) and create the socket in one atomic step:

def open_ephemeral_socket():
    tcp = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    tcp.bind(('', 0))
    return tcp.getsockname(), address

In this case you would then use the returned socket not try to open a new one on the same port.

Note that the atomic nature of this code is backed up by the operating system. It allocates the port and creates the socket in one go so there is no chance of any race condition.

more modern and pythonic:

import socket
from contextlib import closing


def get_local_ip() -> str:
    with closing(socket.socket(socket.AF_INET, socket.SOCK_STREAM)) as s:
        s.connect(("8.8.8.8", 53))
        return s.getsockname()[0]