首先，假设您有MRO链 从最底层的子类init方法开始，任何使用super()方法的类都将跳转到相应的链位置，而任何不使用super()方法的类也将跳转到相应的链位置。

这两种方法都很有效。使用super()的方法为子类带来了更大的灵活性。

在直接调用方法中，C.__init__可以同时调用A.__init__和B.__init__。

当使用super()时，类需要设计为合作多重继承，其中C调用super，它调用A的代码，后者也将调用super，后者调用B的代码。有关使用super可以做什么，请参阅http://rhettinger.wordpress.com/2011/05/26/super-considered-super。

[稍后编辑的回答问题]

所以除非我知道/控制类的初始化，否则我 我不能对我所在的班级做出安全的选择 写作(C)。

参考的文章展示了如何通过在a和b周围添加包装器类来处理这种情况。在“如何合并非合作类”一节中有一个设计好的示例。

有人可能希望多重继承更简单，让你毫不费力地组合Car和Airplane类来获得FlyingCar，但现实是，单独设计的组件通常需要适配器或包装器，然后才能像我们希望的那样无缝地组合在一起:

另一个想法是:如果你对使用多重继承的组合功能不满意，你可以使用组合来完全控制在什么情况下调用哪些方法。

问题的答案取决于一个非常重要的方面:基类是为多重继承设计的吗?

有3种不同的场景:

The base classes are unrelated, standalone classes. If your base classes are separate entities that are capable of functioning independently and they don't know each other, they're not designed for multiple inheritance. Example: class Foo: def __init__(self): self.foo = 'foo' class Bar: def __init__(self, bar): self.bar = bar Important: Notice that neither Foo nor Bar calls super().__init__()! This is why your code didn't work correctly. Because of the way diamond inheritance works in python, classes whose base class is object should not call super().__init__(). As you've noticed, doing so would break multiple inheritance because you end up calling another class's __init__ rather than object.__init__(). (Disclaimer: Avoiding super().__init__() in object-subclasses is my personal recommendation and by no means an agreed-upon consensus in the python community. Some people prefer to use super in every class, arguing that you can always write an adapter if the class doesn't behave as you expect.) This also means that you should never write a class that inherits from object and doesn't have an __init__ method. Not defining a __init__ method at all has the same effect as calling super().__init__(). If your class inherits directly from object, make sure to add an empty constructor like so: class Base(object): def __init__(self): pass Anyway, in this situation, you will have to call each parent constructor manually. There are two ways to do this: Without super class FooBar(Foo, Bar): def __init__(self, bar='bar'): Foo.__init__(self) # explicit calls without super Bar.__init__(self, bar) With super class FooBar(Foo, Bar): def __init__(self, bar='bar'): super().__init__() # this calls all constructors up to Foo super(Foo, self).__init__(bar) # this calls all constructors after Foo up # to Bar Each of these two methods has its own advantages and disadvantages. If you use super, your class will support dependency injection. On the other hand, it's easier to make mistakes. For example if you change the order of Foo and Bar (like class FooBar(Bar, Foo)), you'd have to update the super calls to match. Without super you don't have to worry about this, and the code is much more readable. One of the classes is a mixin. A mixin is a class that's designed to be used with multiple inheritance. This means we don't have to call both parent constructors manually, because the mixin will automatically call the 2nd constructor for us. Since we only have to call a single constructor this time, we can do so with super to avoid having to hard-code the parent class's name. Example: class FooMixin: def __init__(self, *args, **kwargs): super().__init__(*args, **kwargs) # forwards all unused arguments self.foo = 'foo' class Bar: def __init__(self, bar): self.bar = bar class FooBar(FooMixin, Bar): def __init__(self, bar='bar'): super().__init__(bar) # a single call is enough to invoke # all parent constructors # NOTE: `FooMixin.__init__(self, bar)` would also work, but isn't # recommended because we don't want to hard-code the parent class. The important details here are: The mixin calls super().__init__() and passes through any arguments it receives. The subclass inherits from the mixin first: class FooBar(FooMixin, Bar). If the order of the base classes is wrong, the mixin's constructor will never be called. All base classes are designed for cooperative inheritance. Classes designed for cooperative inheritance are a lot like mixins: They pass through all unused arguments to the next class. Like before, we just have to call super().__init__() and all parent constructors will be chain-called. Example: class CoopFoo: def __init__(self, **kwargs): super().__init__(**kwargs) # forwards all unused arguments self.foo = 'foo' class CoopBar: def __init__(self, bar, **kwargs): super().__init__(**kwargs) # forwards all unused arguments self.bar = bar class CoopFooBar(CoopFoo, CoopBar): def __init__(self, bar='bar'): super().__init__(bar=bar) # pass all arguments on as keyword # arguments to avoid problems with # positional arguments and the order # of the parent classes In this case, the order of the parent classes doesn't matter. We might as well inherit from CoopBar first, and the code would still work the same. But that's only true because all arguments are passed as keyword arguments. Using positional arguments would make it easy to get the order of the arguments wrong, so it's customary for cooperative classes to accept only keyword arguments. This is also an exception to the rule I mentioned earlier: Both CoopFoo and CoopBar inherit from object, but they still call super().__init__(). If they didn't, there would be no cooperative inheritance.

底线:正确的实现取决于继承的类。

构造函数是类的公共接口的一部分。如果类被设计为mixin或用于合作继承，则必须将其记录在案。如果文档中没有提及此类内容，那么可以安全地假设该类不是为合作多重继承而设计的。

如果你有多个来自第三方库的子类，那么不，没有盲目的方法来调用基类__init__方法(或任何其他方法)，不管基类是如何编程的，它们实际上都是有效的。

Super使编写用于协作实现方法的类成为复杂的多重继承树的一部分成为可能，类作者不需要知道这些继承树。但是没有办法使用它来正确地从可能使用或不使用super的任意类继承。

Essentially, whether a class is designed to be sub-classed using super or with direct calls to the base class is a property which is part of the class' "public interface", and it should be documented as such. If you're using third-party libraries in the way that the library author expected and the library has reasonable documentation, it would normally tell you what you are required to do to subclass particular things. If not, then you'll have to look at the source code for the classes you're sub-classing and see what their base-class-invocation convention is. If you're combining multiple classes from one or more third-party libraries in a way that the library authors didn't expect, then it may not be possible to consistently invoke super-class methods at all; if class A is part of a hierarchy using super and class B is part of a hierarchy that doesn't use super, then neither option is guaranteed to work. You'll have to figure out a strategy that happens to work for each particular case.

如果您可以控制A和b的源代码，那么任何一种方法(“新样式”或“旧样式”)都可以工作。否则，可能需要使用适配器类。

源代码可访问:正确使用“new style”

class A(object):
    def __init__(self):
        print("-> A")
        super(A, self).__init__()
        print("<- A")

class B(object):
    def __init__(self):
        print("-> B")
        super(B, self).__init__()
        print("<- B")

class C(A, B):
    def __init__(self):
        print("-> C")
        # Use super here, instead of explicit calls to __init__
        super(C, self).__init__()
        print("<- C")

>>> C()
-> C
-> A
-> B
<- B
<- A
<- C

这里，方法解析顺序(MRO)规定如下:

C(A, B)先指示A，然后B。MRO是C -> A -> B ->对象。 super(A, self).__init__()沿着在C.__init__到B.__init__中初始化的MRO链继续。 super(B, self).__init__()沿着在C.__init__中初始化的MRO链继续到object.__init__。

可以说，这个案例是为多重继承而设计的。

源代码可访问:正确使用“旧样式”

class A(object):
    def __init__(self):
        print("-> A")
        print("<- A")

class B(object):
    def __init__(self):
        print("-> B")
        # Don't use super here.
        print("<- B")

class C(A, B):
    def __init__(self):
        print("-> C")
        A.__init__(self)
        B.__init__(self)
        print("<- C")

>>> C()
-> C
-> A
<- A
-> B
<- B
<- C

这里，MRO无关紧要，因为显式地调用了A.__init__和B.__init__。C类(B, A):同样有效。

尽管这种情况不像前一种情况那样是为新样式中的多重继承而“设计”的，但多重继承仍然是可能的。

现在，如果A和B来自第三方库怎么办——也就是说，您无法控制A和B的源代码?简单的回答是:您必须设计一个适配器类来实现必要的super调用，然后使用一个空类来定义MRO(参见Raymond Hettinger关于super的文章——特别是“如何合并非协作类”一节)。

第三方父级:A不实现超级;B确实

class A(object):
    def __init__(self):
        print("-> A")
        print("<- A")

class B(object):
    def __init__(self):
        print("-> B")
        super(B, self).__init__()
        print("<- B")

class Adapter(object):
    def __init__(self):
        print("-> C")
        A.__init__(self)
        super(Adapter, self).__init__()
        print("<- C")

class C(Adapter, B):
    pass

>>> C()
-> C
-> A
<- A
-> B
<- B
<- C

类Adapter实现了super，这样C就可以定义MRO，当super(Adapter, self).__init__()被执行时，MRO就开始发挥作用了。

如果反过来呢?

第三方父类:A实现super;B没有

class A(object):
    def __init__(self):
        print("-> A")
        super(A, self).__init__()
        print("<- A")

class B(object):
    def __init__(self):
        print("-> B")
        print("<- B")

class Adapter(object):
    def __init__(self):
        print("-> C")
        super(Adapter, self).__init__()
        B.__init__(self)
        print("<- C")

class C(Adapter, A):
    pass

>>> C()
-> C
-> A
<- A
-> B
<- B
<- C

这里的模式相同，只是在Adapter.__init__中执行顺序发生了切换;首先是超级调用，然后是显式调用。注意，每个带有第三方父类的情况都需要一个唯一的适配器类。

因此，似乎除非我知道/控制从(A和B)继承的类的初始化，否则我无法为我正在编写的类(C)做出安全的选择。

虽然您可以通过使用适配器类来处理无法控制A和B的源代码的情况，但确实必须知道父类的init是如何实现super的(如果有的话)。

它遵循MRO规则，并调用init。