I want to call parent class method which is overridden in child class through child class object in Python

Question

class abc():
    def xyz(self):
        print("Class abc")

class foo(abc):
    def xyz(self):
        print("class foo")

x = foo()

I want to call xyz() of the parent class, something like;

x.super().xyz()

Answer 1

With single inheritance like this it's easiest in my opinion to call the method through the class, and pass self explicitly:

abc.xyz(x)

Using super to be more generic this would become (though I cannot think of a good use case):

super(type(x), x).xyz()

Which returns a super object that can be thought of as the parent class but with the child as self.

If you want something exactly like your syntax, just provide a super method for your class (your abc class, so everyone inheriting will have it):

def super(self):
    return super(type(self), self)

and now x.super().xyz() will work. It will break though if you make a class inheriting from foo, since you will only be able to go one level up (i.e. back to foo).

There is no "through the object" way I know of to access hidden methods.

Just for kicks, here is a more robust version allowing chaining super calls using a dedicated class keeping tracks of super calls:

class Super:
    def __init__(self, obj, counter=0):
        self.obj = obj
        self.counter = counter

    def super(self):
        return Super(self.obj, self.counter+1)

    def __getattr__(self, att):
        return getattr(super(type(self.obj).mro()[self.counter], self.obj), att)


class abc():
    def xyz(self):
        print("Class abc", type(self))

    def super(self):
        return Super(self)


class foo(abc):
    def xyz(self):
        print("class foo")


class buzz(foo):
    def xyz(self):
        print("class buzz")


buzz().super().xyz()
buzz().super().super().xyz()

results in

class foo
Class abc