Get the defining class for a unbound method in Python

method_reflection.py

def get_class(method):
    """
    Get the class that the defined the provided method.
    In Python 3 the concept of unbound method is removed and raw method
    references are just plain functions. But it is sometimes useful to get the
    class that defined that method. Unfortunately that turns out to be hard.
    But there is a sure fire way of doing it, because Python exposes the
    variables that a method can access, which by definition includes the
    surronding class, so just go through that and find the class.
    Of course it's not quite that simple. First off to find the class we check
    if the value has an attribute that is named the same as the method, and is
    a reference to the same. But we also have to check the qualifed name to
    avoid subclasses shadowing the origin class.
    >>> class Foo:
    ...     def bar(self):
    ...         pass
    ...
    >>> get_class(Foo.bar) is Foo
    True

    >>> class Foo:
    ...     def bar(self):
    ...         pass
    ...
    >>> class Bar(Foo):
    ...     pass # don't override bar, it's still Foo's version
    ...
    >>> get_class(Foo.bar) is Foo
    True
    >>> get_class(Bar.bar) is Foo
    True

    """
    for value in method.__globals__.copy().values():
        if hasattr(value, method.__name__) and \
           getattr(value, method.__name__) is method and \
           f"{value.__qualname__}.{method.__name__}" == method.__qualname__:

            return value