Can Python select what network adapter when opening a socket?

I can't speak much for Windows, but on Linux the interface is normally not chosen until a routing decision is made, therefore you usually don't have a say on which interface your packets leave.

You do have the option though, of using SO_BINDTODEVICE (see man 7 socket) on Linux. This binds a socket to a device, however, only root can set this option on a socket.

Just checked, and the python socket library doesn't have SO_BINDTODEVICE defined, but you get it from socket.h:

# from socket.h
# define SO_BINDTODEVICE 25

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.setsockopt(socket.SOL_SOCKET, 25, 'eth0')

See also:

• Linux: How do I force a specific network interface to be used?
• How do I make an outgoing socket to a SPECIFIC network interface?

On Windows, if you know the IP address of the interface you want to use, just bind to that before you connect. On Linux,use socket option SO_BINDTODEVICE as suggested by JimB (seems to be a privileged call too).

i.e. on Windows

import socket
s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.bind(('192.168.0.1', 0))
s.connect(('...'))

Binding source address under Windows, selects the interface with the same IP address as that device, even if that IP address has a higher routing metric cost. This doesn't work under Linux though, as it always overwrites the source address with the IP address of the selected device. Routing is done based solely on the destination address. The only exception it seems is if you set source address to 127.0.0.1, then Linux prevents these packets from going out of that box.