如何使用Python的super()更新父值? [英] How can I use Python's super() to update a parent value?

问题描述

我是继承的新手，以前关于继承和Python的super()函数的所有讨论都让我有些头疼.我目前使用以下代码更新父对象的值.

I'm new to inheritance and all of the previous discussions about inheritance and Python's super() function are a bit over my head. I currently use the following code to update a parent object's value.

#!/usr/bin/env python
# test.py

class Master(object):
    mydata = []
    def __init__(self):
        s1 = Sub1(self)
        s2 = Sub2(self)

class Sub1(object):
    def __init__(self,p):
        self.p = p
        self.p.mydata.append(1)

class Sub2(object):
    def __init__(self,p):
        self.p = p
        self.p.mydata.append(2)

if __name__ == "__main__":
    m = Master()
    print m.mydata

此命令行返回如下:

user @ host:〜$ ./test.py
[1、2]

user@host:~$ ./test.py
[1, 2]

是否有更好的方法通过super()而不是将"self"引用传递给孩子?

Is there a better way to do this with super() instead of passing the the "self" reference to the child?

推荐答案

super仅适用于类继承结构，其中Sub1和Sub2是Master的子类.

super only applies to class inheritance structures, where Sub1 and Sub2 are subclasses of Master.

在您的示例中，您使用一个包含结构，Sub1和Sub2是Master的属性，并且对super调用没有用.

In your example, you use a containment structure, Sub1 and Sub2 are attributes of Master, and you have no use for super calls.

此外，您通常真的不想使用可变列表作为类属性；追加它会全局(而不是按实例)更改列表的一个副本(在类中定义)；而是使用Master.__init__方法初始化列表:

Also, you generally really do not want to use a mutable list as a class attribute; appending to it will alter the one copy of the list (defined in the class) globally, not per instance; initiate the list in the Master.__init__ method instead:

class Master(object):
    mydata = None
    def __init__(self):
        self.mydata = []

调用__init__函数来设置新实例，并通过在其中为self分配新的空列表来确保每个实例都有自己的副本.

The __init__ function is called to set up a new instance, and by assigning a new empty list to self there, you ensure that each instance has it's own copy.

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