我有一个孩子类，两个父母的函数名称相同。

我需要根据构造函数（__init__）决定（基于要创建的对象的类型）哪个子类的对象需要调用哪个父类的函数。有没有一种方法可以将默认函数调用动态链接/切换到父对象之一。下面给出了示例代码。我不想使用对象容器，因为它很难维护和重构。

class Square: 
 def perform(self, num):
     print(num * num)

class Cube: 
 def perform(self, num):
     print(num * num * num)

class Exponent(Square, Cube):
 def __init__(self, type):
    if type == 'square':
        # inherit default functions from square <--- ??
    else:
        # inherit default functions from cube <--- ??

    square = Square()  # <-- i don't want to do this
    cube = Cube()      # <-- i don't want to do this

Exponent('cube').perform(2)    # --> should print '8'
Exponent('square').perform(2)  # --> should print '4'

Exponent('cube').cube.perform(2)      # <-- i don't want to do this
Exponent('square').square.perform(2)  # <-- i don't want to do this

下面给出一种方法，但是涉及到复制所有父类函数，这太过分了：

class a:
    def x (self):
        print('a')

class b:
    def x (self):
        print('b')

class c(a,b):
    def __init__(self, type_):
        if type_ == 'a':
            self.ref = a
        else:
            self.ref = b

    def x(self):
        self.ref.x(self)


c('a').x()
c('b').x()