Forcing overriden method to call base method implementation

Question

class Base:
    def method(self):
        print("Base method called")

class Derived(Base):
    def method(self):
        print("Derived method called")

I would like to find a way for the derived method to call the base method implementation automatically,or better yet, to force the calling of the base method before the derived method is called.

ex:

derived = Derived()
derived.method()

desired output:

Base method called
Derived method called

I wanted to avoid explicitly calling the base method inside the derived method.

I tried messing with __init_subclass__ hacks:

class Base:
    def __init_sublcass__(cls, *args, **kwargs):
        def wrapper(self):
            super().method()
            self.method()
        cls.method = wrapper
    
    def method(self):
        print("Base method called")

It works but pylance has problems with it and it's not really ideal or easy to understand.

Answer 1

I think you might force the method() to always call Base.method() using a metaclass.

How about something like:

class Base_Meta(type):
    def __init__(cls, name, bases, dct):
        def method_new(self):
            Base.method(self)
            cls.method_old(self)

        if cls.__name__ != "Base":
            cls.method_old = cls.method
            cls.method = method_new

class Base(metaclass=Base_Meta):
    def method(self):
        print("Base method called")

class Derived(Base):
    def method(self):
        print("Derived method called")

d = Derived()
d.method()

I think that will give you:

Base method called
Derived method called