What does it mean when a parent class has a super()?

问题

As far as i know, super() is used to call the methods overwritten in the subclass (like an overwritten __init__ in the subclass).

Also, i know that all python classes inherit from a special class called object.

While studying OOP i ran into a weird example:

class A(object):
    def __init__(self):
        print('A.__init__')
        super().__init__()

class B(object):
    def __init__(self):
        print('B.__init__')
        super().__init__()

class C(A, B):
    def __init__(self):
        print('C.__init__')
        super().__init__()

I understand most of this except...why does class A has a super().__init__()? It's the parent class, it shouldn't need to inherit from anyone. And the special object class doesn't do anything so i don't know why it would need it.

This is not the only case i have seen a super in the parent class, in a previous question i made, a solution to aproblem involved it but i didn't understood it. I also can't find an easy explanation of this usage of super() inheritance.

In simple terms...why would i need to use super() in a parent class?

回答1:

Consider the following example:

class A:
    def __init__(self, a, **kwargs):
        print('A.__init__ called')
        self.a = a
        super().__init__(**kwargs)
        print('A.__init__ finished')

class B:
    def __init__(self, b, **kwargs):
        print('B.__init__ called')
        self.b = b
        super().__init__(**kwargs)
        print('B.__init__ finished')

class C(A, B):
    def __init__(self, c, **kwargs):
        print('C.__init__ called')
        self.c = c
        super().__init__(**kwargs)
        print('C.__init__ finished')

When you create a C object, the output is as follows:

>>> C(a=1, b=2, c=3)
C.__init__ called
A.__init__ called
B.__init__ called
B.__init__ finished
A.__init__ finished
C.__init__ finished
<__main__.C object at 0x7fd15abcab70>

From the order that the __init__ methods are called, and the fact that they finish in reverse order, we can see that C.__init__ calls A.__init__, which calls B.__init__. That is, although A has no explicit parent class (so its direct parent is object), the super().__init__ call in A actually calls B.__init__.

This is because B is the next class after A in C's method resolution order (MRO). The A class needs to call super() because although its superclass is object, Python allows multiple inheritance, so A cannot guarantee that it is the last class before object in the MRO.

>>> C.__mro__
(<class '__main__.C'>, <class '__main__.A'>, <class '__main__.B'>, <class 'object'>)

See Raymond Hettinger's article Python's super() considered super! for a more detailed explanation about the MRO and "cooperative multiple inheritance techniques".

回答2:

In simple terms...why would i need to use super() in a parent class?

You don't. Unless the parent class is itself a child of another higher class there is no reason to call super() in the parent class and it looks like the tutorial you're reading might have accidentally just copy/pasted some additional code.

On another note, you also don't need the () brackets to define a class and furthermore unless you're working in Python2 (which you shouldn't be in 2020 as Python2.x is no longer supported from 2020) you don't need to add (object) behind a class name, as that was purely a way for classic Python2 classes to take on the new style format, but in Python3 all classes are new style classes.

So, the correct way for your code above to appear would be the following:

class A:
    def __init__(self):
        print('A.__init__')
        # you don't need a super() here as class A isn't inheriting anything

class B:
    def __init__(self):
        print('B.__init__')
        # you don't need the super() here either as B also isn't inheriting anything

class C(A, B):
    def __init__(self):
        print('C.__init__')
        super().__init__()

来源：https://stackoverflow.com/questions/59832892/what-does-it-mean-when-a-parent-class-has-a-super

标签

python

oop

super