如何在python中创建IPv6套接字？为什么会出现socket.error：（22，'Invalid arguments'）？
问题内容：
我想在python上创建Ipv6套接字，我这样做是这样的：
#!/usr/bin/env python
import sys
import struct
import socket
host = 'fe80::225:b3ff:fe26:576'
sa = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM)
sa.bind((host , 50000))
但是失败了：
socket.error: (22, 'Invalid argument') ?
谁能帮我？谢谢！
我像这样重做，但仍然无法正常工作
>>>host = 'fe80::225:b3ff:fe26:576'
>>>sa = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM)
>>>res = socket.getaddrinfo(host, port, socket.AF_UNSPEC, socket.SOCK_DGRAM, 0, socket.AI_PASSIVE)
>>>family, socktype, proto, canonname, sockaddr = res[0]
>>>print sockaddr
('fe80::225:b3ff:fe26:576', 50001, 0, 0)
>>>sa.bind(sockaddr)
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<string>", line 1, in bind
socket.error: (22, 'Invalid argument')
问题答案：
这个问题有两个部分
首要问题
您应该使用sa.bind（sockaddr），其中从getaddrinfo获得sockaddr
>>> HOST = 'localhost'
>>> PORT = 50007
>>> res = socket.getaddrinfo(HOST, PORT, socket.AF_UNSPEC, socket.SOCK_DGRAM, 0, socket.AI_PASSIVE)
>>> family, socktype, proto, canonname, sockaddr = res[1]
>>> proto
17
>>> sockaddr
('fe80::1%lo0', 50007, 0, 1)
第二期
如果您查看套接字文档中提供的示例，请访问
套接字接受三个参数
socket( [family[, type[, proto]]])
根据文档
Create a new socket using the given address family,
socket type and protocol number. The address family
should be AF_INET (the default), AF_INET6 or AF_UNIX.
The socket type should be SOCK_STREAM (the default),
SOCK_DGRAM or perhaps one of the other "SOCK_" constants.
The protocol number is usually zero and may be omitted in that case.
而且，如果您使用getaddressinfo获取proto的值，则该值不同于默认值0
但是，当我执行以下命令时，会得到一个不同的协议值-17。
当然，socket.has_ipv6对我来说是正确的。
