On Friday, May 3, 2013 10:23:38 PM UTC-4, Chris Angelico wrote: > On Sat, May 4, 2013 at 12:13 PM, Pedro <pe...@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(). > > > > > 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 > > connections. You can bypass this check by setting the SO_REUSEADDR > > option - lemme hunt that down in the Python docs, haven't done that in > > Python for a while... > > > > http://docs.python.org/3.3/library/socket.html#socket.socket.setsockopt > > > > s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1) > > > > That should do the job. > > > > ChrisA
Thanks Chris, can you elaborate on the accept() call should block? -- http://mail.python.org/mailman/listinfo/python-list
