why is there now execption for windows? trying to listen twice tothe same port
in [Python]
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From: News123 on 20 Mar 2010 09:38 I'm having a small multiprocessing manager: # ########################## import socket,sys from multiprocessing.managers import BaseManager mngr = BaseManager(address=('127.0.0.1',8089),authkey='verysecret') try: srvr = mngr.get_server() except socket.error as e: print "probably address already in use" sys.exit() print "serving" srvr.serve_forever() Under linux this script can only be run once. The second call will raise an exception, as the previous program is already listening to pot 8089. Under Windows however the program can be started twice. and will print twice "serving". This surprises me Howver only one of them will successfully listen and respond to connections, but I don't get an exception. Is this to be expected? I'd like to be sure, that the manager is only started once and that it can be started anytime it's not up. N |