2013-03-26 93 views
8

我正在學習網絡編程,並嘗試通過此示例掌握套接字的基礎知識。Socket.error:無效提供的參數

import socket,sys 


s = socket.socket(socket.AF_INET,socket.SOCK_DGRAM) 

MAX = 65535 
PORT = 1060 

if sys.argv[1:] == ['server']: 
    s.bind(('127.0.0.1',PORT)) 
    print 'Listening at ' , s.getsockname() 
    while True: 
     data,address = s.recvfrom(MAX) 
     print ' The address at ' , address , ' says ' , repr(data) 
     s.sendto('your data was %d bytes' % len(data),address) 

elif sys.argv[1:] == ['client']: 
    print ' Address before sending ' ,s.getsockname() 
    s.sendto('This is the message',('127.0.0.1',PORT)) 
    print ' Address after sending ' ,s.getsockname() 
    data,address = s.recvfrom(MAX) 
    print ' The server at ' , address , ' says ' , repr(data) 

else: 
    print >> sys.stderr, 'usage: udp_local.py server | client ' 

然而,它扔了一個異常說通過getsockname給出的參數()無效專門就行22.The代碼是正確的,因爲據我know.Here是例外

Traceback (most recent call last): 
    File "udp_local.py", line 23, in <module> 
    print ' Address before sending ' ,s.getsockname() 
    File "c:\Python27\lib\socket.py", line 224, in meth 
    return getattr(self._sock,name)(*args) 
error: [Errno 10022] An invalid argument was supplied 

使用PyScripter 2.5.3.0 x86

+0

那是Winsock的? – wRAR 2013-03-26 13:55:16

+0

一個套接字可能實際上不會在發送之前擁有一個地址,除非您先調用bind。在Mac上,我沒有收到錯誤,但返回的端口是'0'(表示尚未分配端口)。 – robertklep 2013-03-26 14:07:56

+0

我正在使用Python.The標準套接字模塊 – devsaw 2013-03-26 14:27:53

回答

8

那麼我得到了問題。套接字沒有地址,直到其綁定或數據發送。 只是要評論它。

elif sys.argv[1:] == ['client']: 
## print ' Address before sending ' ,s.getsockname() 

感謝

+1

你應該添加一個鏈接並接受你的答案 – 2014-10-01 13:47:37