How to inspect a method without instance?

Question

I need to:

• Receive a method as argument (directly from the class, without instance)
• Create an instance
• Execute said method from instance

The thing is, how can I reliably get the class out of the method? I've tried researching something in the inspect module, but because there is no instance, it thinks of the method as a function rather than a method.

Here's some example of what I'm trying to do:

def execute_method(method):
    cls = get_class_from_method(method)
    obj = cls()
    getattr(obj, method.__name__)()


def get_class_from_method(method):
    pass # how?


execute_method(HelloView.say_hello)

Answer 1

A Python 2 method objects have a im_class attribute referencing the class of the instance they are bound to:

cls = method.im_class

For an unbound method, in Python 3, you'll be out of luck (the function is returned entirely unchanged, with no reference to the class). Python 2 returns an unbound method type with the im_class attribute still there:

>>> class Foo(object):
...     def bar(self): pass
...
>>> Foo.bar.im_class
<class '__main__.Foo'>

In Python 3, your options are far more limited and are error-prone. You could look at the __qualname__ attribute of the function, and deduce from that what the class might be bound to in the global namespace:

>>> class Foo:
...     def bar(self): pass
...
>>> Foo.bar
<function Foo.bar at 0x10e97b268>
>>> Foo.bar.__qualname__
'Foo.bar'
>>> Foo.bar.__qualname__.rpartition('.')[0]
'Foo'
>>> Foo.bar.__globals__.get(Foo.bar.__qualname__.rpartition('.')[0])
<class '__main__.Foo'>

However, if the class was created in a function (and is thus a local), or the class was replaced via a class decorator or simply renamed in the global namespace, the above trick at best won't work, at worst give you an entirely different object, and you can't know if this is the case.