假设我有一个多重继承的场景:
class A(object):
# code for A here
class B(object):
# code for B here
class C(A, B):
def __init__(self):
# What's the right code to write here to ensure
# A.__init__ and B.__init__ get called?
有两种典型的方法来编写C语言的__init__:
(老式)ParentClass.__init__(自我)
(new -style) super(DerivedClass, self).__init__()
然而，在任何一种情况下，如果父类(A和B)不遵循相同的约定，那么代码将不能正确工作(有些可能会丢失，或被多次调用)。
正确的方法是什么来着?说“保持一致，遵循其中一个”很容易，但如果A或B来自第三方库，那怎么办?是否有一种方法可以确保所有父类构造函数都被调用(并且以正确的顺序，并且只调用一次)?
编辑:看看我的意思，如果我这样做:
class A(object):
def __init__(self):
print("Entering A")
super(A, self).__init__()
print("Leaving A")
class B(object):
def __init__(self):
print("Entering B")
super(B, self).__init__()
print("Leaving B")
class C(A, B):
def __init__(self):
print("Entering C")
A.__init__(self)
B.__init__(self)
print("Leaving C")
然后我得到:
Entering C
Entering A
Entering B
Leaving B
Leaving A
Entering B
Leaving B
Leaving C
注意B的init被调用了两次。如果我这样做:
class A(object):
def __init__(self):
print("Entering A")
print("Leaving A")
class B(object):
def __init__(self):
print("Entering B")
super(B, self).__init__()
print("Leaving B")
class C(A, B):
def __init__(self):
print("Entering C")
super(C, self).__init__()
print("Leaving C")
然后我得到:
Entering C
Entering A
Leaving A
Leaving C
注意B的init从未被调用。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。
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。
下面是我如何在Python 3中使用super()实现多重继承
class A:
def __init__(self, a, b, **kwargs):
print("Class A initiallised")
self.a = a
self.b = b
super().__init__(**kwargs)
print("Class A initiallisation done")
def __str__(self):
return f"{self.a} and {self.b}"
def display_a(self):
return f"{self.a} and {self.b}"
class C:
def __init__(self, c, d, **kwargs):
print("Class C initiallised")
self.c = c
self.d = d
super().__init__(**kwargs)
print("class c initiallisation done")
def __str__(self):
return f"{self.c} and {self.d}"
def display_c(self):
return f"{self.c} and {self.d}"
class D(A,C):
def __init__(self, e, **kwargs):
print("Class D initiallised")
super().__init__(**kwargs)
self.e = e
print("Class D initiallisation done")
def __str__(self):
return f"{self.e} is e,{self.b} is b,{self.a} is a,{self.d} is d,{self.c} is c"
if __name__ == "__main__":
d = D(a=12, b=13, c=14, d=15, e=16)
print(d)
d.display_c()
d.display_a()
