Python的socket.connect奇怪的行为

Question

我不能google一下，并不能明白为什么这个代码工作它的工作原理

Python 3.5.2 (default, Sep 28 2016, 18:08:09) 
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.38)] on darwin 
Type "help", "copyright", "credits" or "license" for more information. 
>>> import socket 
>>> s = socket.socket(socket.SOCK_DGRAM) 
>>> s.connect(('127.0.0.1', 33000)) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
ConnectionRefusedError: [Errno 61] Connection refused 
>>> s.family 
<AddressFamily.AF_INET: 2> 
>>> 
>>> 
>>> s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM) 
>>> s.family 
<AddressFamily.AF_INET: 2> 
>>> s.connect(('127.0.0.1', 33000)) 
>>> 

编辑的方式：

Python 3.5.2 (default, Sep 28 2016, 18:08:09) 
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.38)] on darwin 
Type "help", "copyright", "credits" or "license" for more information. 
>>> s2 = socket.socket(socket.SOCK_DGRAM) 
KeyboardInterrupt 
>>> import socket 
>>> s = socket.socket(socket.SOCK_DGRAM) 
>>> s.connect(('127.0.0.1', 33000)) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
ConnectionRefusedError: [Errno 61] Connection refused 
>>> s.family 
<AddressFamily.AF_INET: 2> 
>>> s.type 
<SocketKind.SOCK_STREAM: 1> 
>>> 

貌似这个是TCP套接字=）我猜Python做了一些检查并丢弃了不正确的值，但没有提供任何有关该技巧的信息。

Answer 1

从documentation：

socket.socket([family[, type[, proto]]]) 

随着socket.socket(socket.SOCK_DGRAM)使用SOCK_DGRAM家庭即使这是一个类型。这会导致奇怪的问题，如你看到的那个。