How to create IPv6 socket at python? Why got the socket.error: (22, 'Invalid argument')?

Question

I want to ceate Ipv6 socket at python, I do it like this:

#!/usr/bin/env python
import sys
import struct
import socket

host = \'fe80::225:b3ff:fe26:576\'
sa         


        

Answer 1

The other thing missing from the above is that fe80::* are link-local addresses.As such you can't technically use them without a Scope ID.
Just in case that didn't sink in, "link-local" means that the address can only be used on a specific link, but the corollary is that since the address is link-specific theoretically you could have that same address on more than one link.
Now granted that in most cases a link-local address is generated from the already unique hardware (EUI-48) address by converting it to an EUI-64 and then flipping on the U/L bit (see) but this is not the only way you can generate a link-local address and other mechanisms could result in re-using the address (I don't think there's anything that prohibits this). So now you know why it didn't work.
The next question is how do you fix it, or rather where do you get the Scope ID. This is unfortunately non-obvious. If you read the IPV6 RFCs that discuss "Scope" you find that it is defined in a general fashion. In practice if you are running this code in Linux (maybe also Windows/Mac?) you can use the interface index.

Note that there's currently a bug in nss-mdns such that .local names are returned with scope id always zero so in Linux using .local names basically doesn't work unless your code has already figured out the scope id and sets it by itself.