python: call a parent class' recursive function

Question

I cannot quite understand the following result:

class A:
    def a(self, i):
        if i <= 0:
            return
        print("A", i)
        self.a(i - 1)  # line 6


class B(A):
    def a(self, i):
        print("B", i)
        super().a(i)   # line 12


if __name__ == '__main__':
    b = B()
    b.a(3)

result is:

B 3
A 3
B 2
A 2
B 1
A 1
B 0

In line 12, it calls the parent class A's function, however when a() recursively call itself, it uses the B's version. Why it happens?

How can I get the following result (I still want to override parent's function a()):

B 3
A 3
A 2
A 1

I want to force the instance only uses function in parent's version.

I have this requirement since I meet similar problem in some practical problem. I still have to name the function as a in class B. However most part of logic is duplicate in A and I want to reuse it.

Right now I can only use the following way to implement B:

class B(A):
    def a(self, i):
        print("B", i)
        self.a_helper(i) 

    def a_helper(self, i):
        if i <= 0:
            return
        print("A", i)
        self.a_helper(i - 1)

Methods are looked up based on the class of the object, not the class where the method is defined. Since self is a B instance, self.a() calls the B method. — Barmar
– Barmar, Commented Mar 8, 2022 at 20:19
Don't override method a in class B. Perhaps call the method a different name. — quamrana
– quamrana, Commented Mar 8, 2022 at 20:23
But this negates some of the benefits of OOP. Subclasses are supposed to be able to override the parent methods. — Barmar
– Barmar, Commented Mar 8, 2022 at 20:27

Barmar · Accepted Answer · 2022-03-08 20:28:15Z

1

Call the method directly as an ordinary function, rather than going through the instance.

class A:
    def a(self, i):
        if i <= 0:
            return
        print("A", i)
        A.a(self, i - 1)  # line 6

answered Mar 8, 2022 at 20:28

Barmar

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