you need to pass a service class, not an instance.
see
https://groups.google.com/forum/?hl=en_US&fromgroups=#!searchin/rpyc/service$20arguments/rpyc/AMwdklcRzFA/S7-mf1oqiH4J


-tomer




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*Tomer Filiba*
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On Wed, May 8, 2013 at 8:42 AM, Mark <[email protected]> wrote:

> I am trying to create my own Service using python 2.7 and RPyC 3.2.3.
> I have removed any customl functions and create a service using the
> following syntax:
>
> class MyService(rpyc.Service):
>       # stripped out all my functions for now
>
> # start the service
> t = ThreadedServer(MyService(args.name), port = int(port)) #passed as a
> parameter
>
> t.start()
>
> #################
>
> I then try to connect to this service simply via:
>
> rpyc.utils.factory.connect('localhost',int(replicaPort))
>
> I have also tried:
>
> rpyc.connect('localhost',int(port),MyService
>
> rpyc.connect('',int(port))
> and a number of other variations, but they all give me back the same error:
>
> TypeError  'MyService' object is not callable.
>
> Does anyone know why this is happening or have any suggestions as to how I
> can troubleshoot this error?  Thanks in advance!
>
> -Mark
>
>
>
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