socket programming

[Pedro]

On Friday, May 3, 2013 10:23:38 PM UTC-4, Chris Angelico wrote:
> On Sat, May 4, 2013 at 12:13 PM, Pedro <pedro at ncf.ca> wrote:
> 
> > First - this code constantly loops around an open socket. Is there a way to use something like an interrupt so I don't have to loop constantly to monitor the socket?
> 
> 
> 
> The accept() call should block. It's not going to spin or anything. If
> 
> you need to monitor multiple sockets, have a look at select().
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> 
> 
> > Second, if any part of the program fails execution (crashes) the port often remains open on my windows machine and the only way to close it that i know of is through task manager or by rebooting the machine. Is there an easy way around this problem ? If I don't close the port the program can't open it again and crashes.
> 
> 
> 
> It remains for a short time to ensure that there's no lurking
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> connections. You can bypass this check by setting the SO_REUSEADDR
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> option - lemme hunt that down in the Python docs, haven't done that in
> 
> Python for a while...
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> 
> 
> http://docs.python.org/3.3/library/socket.html#socket.socket.setsockopt
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> 
> 
> s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
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> 
> 
> That should do the job.
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> 
> 
> ChrisA

Thanks Chris, can you elaborate on the accept() call should block?