Better way to overload a method from two different classes in Python

Question

I have two parent classes I cannot redefine (say, A and B), with methods I need to overload using the exact same way (so as to obtain, say, BetterA and BetterB). I could duplicate the code for both classes, but I am not satisfied with that. In Python 3.6, I thought I can get rid of this using multi inheritance and a provider class. This is what I obtained so far:

# here are the initial classes I cannot edit
class A:
    def __init__(self, a=0):
        self.a = a

class B:
    def __init__(self, b=0):
        self.b = b

# here is the provider class
class Provider:
    def __init__(self, c, *args, **kwargs):
        self.c = c

        # more code here

        self.init_child(*args, **kwargs)

# here are the new classes
class BetterA(Provider, A):
    def init_child(self, *args, **kwargs):
        A.__init__(*args, **kwargs)

class BetterB(Provider, B):
    def init_child(self, *args, **kwargs):
        B.__init__(*args, **kwargs)

if __name__ == '__main__':
    a = BetterA(8, a=10)
    b = BetterB(10, b=8)

This works, but it is not very graceful… Especially, if I want to override more methods (always, in the same way), I have to recall them in BetterA and BetterB.

If there a better way to achieve what I want?

Ry- · Accepted Answer · 2017-09-03 23:16:52Z

2

super() works with multiple inheritance:

# here is the provider class
class Provider:
    def __init__(self, c, *args, **kwargs):
        self.c = c

        # more code here

        super().__init__(*args, **kwargs)

# here are the new classes
class BetterA(Provider, A):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

class BetterB(Provider, B):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

if __name__ == '__main__':
    a = BetterA(8, a=10)
    b = BetterB(10, b=8)

and you don’t need to redefine Better*.__init__ if it’s not doing anything beyond calling the super method:

# here are the new classes
class BetterA(Provider, A):
    pass

class BetterB(Provider, B):
    pass

Another option you might prefer depending on whether you need to implement any more functionality in the Better* types is a function to create a class:

def provided(cls):
    class Provided(cls):
        def __init__(self, c, *args, **kwargs):
            self.c = c

            # more code here

            super().__init__(*args, **kwargs)

    return Provided


BetterA = provided(A)
BetterB = provided(B)

answered Sep 3, 2017 at 23:16

Ry-♦

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2 Comments

Neraste Over a year ago

Wow! I would say this is a "super" solution! But I'm wondering: which init the super at line 8 (first code block) is calling and how?

Ry- Over a year ago

@Neraste: It’s calling A.__init__ or B.__init__ according to BetterA.__mro__ or BetterB.__mro__.

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