I am learning networking programming and trying to grasp the basics of sockets through this example.
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 '
However,its throwing up an exception saying the arguments given by getsockname() were invalid specifically on line 22.The code is correct as far as I know.Here's the exception
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
Using PyScripter 2.5.3.0 x86
Well I got the problem.The socket doesn't have an address untill its either binded or data is sent. Just had to comment it out.
elif sys.argv[1:] == ['client']:
## print ' Address before sending ' ,s.getsockname()
Thanks
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