python - 类范围内的闭包

Question

据我了解，函数和类作用域的行为几乎相同:

>>> def x():
...     a = 123
...     print (locals())
... 
>>> x()
{'a': 123}


>>> class x():
...     a = 123
...     print (locals())
... 
{'a': 123, '__module__': '__main__'}

然而，当我定义一个闭包时，行为是不同的。正如预期的那样，函数只是返回本地绑定(bind):

>>> def x():
...     a = 123
...     t = lambda: a
...     return t
... 
>>> x()()
123

而在一个类中，绑定(bind)似乎丢失了:

>>> class x():
...     a = 123
...     t = lambda self: a
... 
>>> x().t()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in <lambda>
NameError: global name 'a' is not defined

谁能解释这种差异？

Answer 1

类作用域是一个临时作用域，它只在执行类定义主体时存在。结果 dict用于创建类命名空间，__dict__类的。

就类中定义的函数而言，下一个“向上”范围是 class 的范围。定义本身。

以下工作正常:

>>> def foo():
...     spam = 'eggs'
...     class Bar(object):
...         def baz(self): return spam
...     return Bar()
... 
>>> foo().baz()
'eggs'

这记录在 pep 227 :

Names in class scope are not accessible. Names are resolved in the innermost enclosing function scope. If a class definition occurs in a chain of nested scopes, the resolution process skips class definitions.

在 class compound statement documentation :

The class’s suite is then executed in a new execution frame (see section Naming and binding), using a newly created local namespace and the original global namespace. (Usually, the suite contains only function definitions.) When the class’s suite finishes execution, its execution frame is discarded but its local namespace is saved. [4] A class object is then created using the inheritance list for the base classes and the saved local namespace for the attribute dictionary.

强调我的；执行框架是临时范围。