多重继承：派生类仅从一个基类获取属性？

我试图学习Python中多重继承的概念。考虑一个类Derv派生自两个类，Base1 and Base2. Derv仅继承第一个基类的成员：

class Base1:
    def __init__(self):
        self.x=10

class Base2:
    def __init__(self):
        self.y=10

class Derv (Base1, Base2):
    pass

d = Derv()
print (d.__dict__)

结果是{ 'x' : 10 }并颠倒继承顺序仅给出{ 'y' : 10 }.

派生类不应该继承两个基类的属性吗？

我不完全明白为什么会这样，但我可以告诉你如何解决它：

由于某种原因，Python 只调用__init__它的父母之一的方法。但是，这可以解决您的问题：

class Base1:
     def __init__(self):
         super().__init__()
         print('b1')
         self.x=10

 class Base2:
     def __init__(self):
         super().__init__() # This line isn't needed. Still not sure why
         print('b2')
         self.y=10

 class Derv (Base1, Base2):

     def __init__(self):
         super().__init__()

 d = Derv()
 print (d.__dict__)


'b2'
'b1'
{'y': 10, 'x': 10}

更新，添加打印语句实际上可以说明情况。例如，

class Base1:
     def __init__(self):
         print('Before Base1 call to super()')
         super().__init__()
         print('b1')
         self.x=10

 class Base2:
     def __init__(self):
         print('Before Base2 call to super()')
         super().__init__() # No remaining super classes to call
         print('b2')
         self.y=10

 class Derv (Base1, Base2):

     def __init__(self):
         super().__init__()

 d = Derv()
 print (d.__dict__)

'Before Base1 call to super()' # Just before the call to super
'Before Base2 call to super()' # Just before call to super (but there are no more super classes)
'b2' # Calls the remaining super's __init__
'b1' # Finishes Base1 __init__
{'y': 10, 'x': 10}