[英]Python socket.connect strange behaviour
我無法用Google搜尋,也無法理解這段程式碼的運作方式
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.
>>> 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))
>>>
編輯:
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>
>>>
看起來像是TCP套接字=)我想Python會做一些檢查並丟棄不正確的值,但是不提供有關該技巧的任何信息。
從文檔中 :
socket.socket([family[, type[, proto]]])
使用socket.socket(socket.SOCK_DGRAM)
,即使這是一種類型,也可以將SOCK_DGRAM
用作家族。 這將導致您遇到的奇怪問題。
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