来自stackoverflow的一个关于 python 嵌套类的问题(nested class)

http://stackoverflow.com/questions/8775246/nested-classes-in-python

有人在stack overflow上提问关于nested class in Python 的问题：

#------------------------------------class A:    def __init__(self):        self.a = 'a'        print self.aclass B(A):    def __init__(self):        self.b = 'b'        A.a = 'a_b'        print self.b, A.a#------------------------------------class C:    class A:        def __init__(self):            self.a = 'a'            print self.a    class B(A):        def __init__(self):            self.b = 'b'            A.a = 'a_b'            print self.b, A.a#------------------------------------#------------------------------------>>> c1 = A()a>>> c1.a'a'>>> c2 = B()b >>> c2.a, c2.b('a_b', 'b')>>> c3 = C()>>> c4 = c3.A()a>>> c4.a'a'>>> c5 = c3.B()b a_b>>> c5.b'b'>>> c5.aTraceback (most recent call last):  File "", line 1, in AttributeError: B instance has no attribute 'a'

Where is the problem in the code? AND In both cases it seems that when B(A) is initialized A() is not initialized. What is the solution for this issue? Note that the term A.__init__() being called inside B()'s __init__()does not work!

前面的c2.a, c2.b 都可以好好的工作，怎么c5.b就挂了呢？

下面是2个有用的回答：

answer 1：

The code executed in a method runs in the local scope of that method. If you access an object that is not in this scope, Python will look it up in the global/module scope, NOT in the class scope or the scope of any enclosing class!

This means that:

A.a = 'a_b'

inside C.B.__init__ will set the class attribute of the global A class, not C.A as you probably intended. For that you would have to do this:

C.A.a = 'a_b'

Also, Python will not call parent methods if you override them in subclasses. You have to do it yourself.

The scoping rules mean that if you wanted to call the __init__ method of the parent class inside C.B.__init__, it has to look like this:

C.A.__init__(self)

and NOT like this:

A.__init__(self)

which is probably what you've tried.

一个方法中的代码运行时如果发现某个对象在这个方法体中没有，那么它会去全局/模块范围里去找，而不是去它的类或嵌套类。

看到这个答案我突然想到了另一个相似的问题，应该是相似的道理：

x = 1def a():    print x    x = 2    print x

answer 2：

Nested classes seems so unpythonic, even if considered as factories. But to answer your question: There simply is no c5.a (instance of C.B). In the init-method of C.B you add to the CLASS C.A an attribute a, but not to C.B! The class A does already have an attribute a, if instantiated! But the object of class B (and even the class) doesn't!

You must also keep in mind, that __init__ is not an constructor like in C++ or Java! The "real constructor" in python would be __new__. __init__ just initializes the instance of a class!

class A:    c = 'class-attribute'    def __init__(self):        self.i = 'instance-attribute'

So in this example c is a class-attribute, where i is an attribute of the instance.

Even more curios, is your attempt to add an attribute to the baseclass at the moment of the instantiation of the child-class. You are not getting a "late" inheritance-attribute that way. You simply add to the class A an additional attribute, which surprises me to even work. I guess you are using python 3.x?

The reason for this behaviour? Well, i guess it has to do with pythons neat feature that in python definitions areexecuted(AFAIK).

The same reason why:

def method(lst = []):

is almost ever a bad idea. the deafult-parameter gets bound at the moment of the definition and you won't generate a new list-object every-time you call the method, but reusing the same list-object.

这里给出了如何解决问题的答案，就是使用显示的调用，比如：C.A.a。另外也说了代码中存在的问题，就是类B中调用A.a其实不是作者的本意，作者用了未绑定方法。

answer 2最后还提到了不要用可变类型做函数可选参数了。

最后作者Update了正确方法:

class Geometry:    class Curve:        def __init__(self,c=1):            self.c = c                          #curvature parameter            print 'Curvature %g'%self.c            pass                                #some codes    class Line(Curve):        def __init__(self):            Geometry.Curve.__init__(self,0)     #the key point            pass                                #some codesg = Geometry()C = g.Curve(0.5)L = g.Line()