Can't define __call__ within __init__?

[Simon Brunning]

On 10 March 2010 13:12, Neal Becker <ndbecker2 at gmail.com> wrote:
> Want to switch __call__ behavior.  Why doesn't this work?  What is the
> correct way to write this?
>
> class X (object):
>    def __init__(self, i):
>        if i == 0:
>            def __call__ (self):
>                return 0
>        else:
>            def __call_ (self):
>                return 1
>
>
> x = X(0)
>
> x()
> TypeError: 'X' object is not callable

__call__ is in the __init__ method's local namespace - you need to
bind it to the class's namespace instead:

        X.__call__ = __call__

But this probably isn't what you want either, since all instances of X
will share the same method.

What are you trying to do? In your simple example, you'd be much
better off with a single __call__ method. But you knew that.

-- 
Cheers,
Simon B.