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如何在派生类中引用基类的属性?

2019-11-11 04:58:11  阅读:85  来源: 互联网

标签:python


我可以做这个:

class Blah2:
    atttr = 5
    aa = atttr


ob = Blah2
print(ob.aa)

http://ideone.com/pKxMc2

所以我认为我也可以这样做:

class Blah1:
    atttr = 5


class Blah2(Blah1):
    aa = atttr


ob = Blah2
print(ob.aa)

不,我不能:http://ideone.com/6HS1MO

吐出以下错误:

Traceback (most recent call last):
  File "./prog.py", line 5, in <module>
  File "./prog.py", line 6, in Blah2
NameError: name 'atttr' is not defined

为什么不起作用以及如何使其起作用?

解决方法:

类块作用域仅在类定义期间临时存在.在定义了类之后,您将必须通过类对象(即Blah1.atttr)访问该属性.

这记录在execution model部分下.

Class definition blocks and arguments to exec() and eval() are special in the context of name resolution. A class definition is an executable statement that may use and define names. These references follow the normal rules for name resolution with an exception that unbound local variables are looked up in the global namespace. The namespace of the class definition becomes the attribute dictionary of the class. The scope of names defined in a class block is limited to the class block; it does not extend to the code blocks of methods – this includes comprehensions and generator expressions since they are implemented using a function scope.

标签:python
来源: https://codeday.me/bug/20191111/2017384.html

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