How do I call super in a method decorator in Python 3? [duplicate]

问题

This question already has an answer here:

Given a method, how do I return the class it belongs to in Python 3.3 onward? (1 answer)

Closed 5 years ago.

How do I fill in the ????

def ensure_finished(iterator):
    try:
        next(iterator)
    except StopIteration:
        return
    else:
        raise RuntimeError


def derived_generator(method):
    def new_method(self, *args, **kwargs):
        x = method(self, *args, **kwargs)
        y = getattr(super(???, self), method.__name__)\
            (*args, **kwargs)

        for a, b in zip(x, y):
            assert a is None and b is None
            yield

        ensure_finished(x)
        ensure_finished(y)

    return new_method

回答1:

EDIT : This does not work for the reasons mentioned in the comments. I'll leave this here so that the next guy trying to answer doesn't do the same thing (until the real answer shows up).

You should use type(self)

Example I simplified your code a bit, but the essence should still be in there

def derived_generator(method):
    def new_method(self, *args, **kwargs):
        x = method(self, *args, **kwargs)
        y = getattr(super(type(self), self), method.__name__)\
            (*args, **kwargs)

        for a, b in zip(y, x):
            yield a, b

    return new_method

class BaseClass(object):
    def iterator(self):
        return [1, 2, 3]

class ChildClass(BaseClass):
    @derived_generator
    def iterator(self):
        return [4, 5, 6]

a = ChildClass()
for x in a.iterator():
    print(x)

来源：https://stackoverflow.com/questions/25921537/how-do-i-call-super-in-a-method-decorator-in-python-3