[Python-Dev] Dinamically set __call__ method

Ethan Furman ethan at stoneleaf.us
Sun Nov 9 04:27:41 CET 2014


On 11/08/2014 02:31 PM, Gregory Ewing wrote:
>
> Seems to depend on how you get hold of the object you're
> inspecting the signature of. I did an experiment:
>
>    class C(object):
>
>    @property
>      def __call__(self):
>        return self.call
>
>    def f(x, y):
>      print("Called f with %s, %s" % (x, y))
>
>    c = C()
>    c.call = f
>    c(17, 42)
>
> prints:
>    Called f with 17, 42
>
> This works too:
>
>    import inspect
>    print(inspect.getargspec(c.__call__))
>
> prints:
>    ArgSpec(args=['x', 'y'], varargs=None, keywords=None, defaults=None)
>
> But
>
>    print(inspect.getargspec(c))
>
> doesn't work:
>
>    TypeError: <__main__.C object at 0x479250> is not a Python function
>
> (BTW, I'm actually surprised that this technique makes c callable.
> There must be more going on that just "look up __call__ in the class
> object", because evaluating C.__call__ just returns the descriptor
> and doesn't invoking the descriptor mechanism.)

Looks like you found a bug in inspect.getargspec.

And the thing going on is the normal python behavior (in __getattribute__, I believe) of examining the returned 
attribute to see if it is a descriptor, and if so invoking it.

--
~Ethan~



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