exec不从闭包中接收变量 [英] exec doesn't pick up variables from closure

问题描述

我很好奇为什么下面的代码引起了 NameError 。

 >>>> s =
 ... foo = [1,2,3] 
 ... def bar（）：
 ... return foo [1] 
。 ..
>>>> namespace = {} 
>>>> exec（s，{'__builtins__'：None}，命名空间）
>>> print namespace 
 {'foo'：[1，2，3]，'bar'：< function bar at 0x7f79871bd0c8>} 
>>命名空间['bar']（）
  

在正常解释器级别， c> foo 在 bar.func_globals 或 bar.func_closure 我想我想知道为什么命名空间['bar'] 不把 foo 在 func_closure ...

解决方案

原来，答案一直在< a href =http://docs.python.org/2/reference/simple_stmts.html#the-exec-statement =nofollow>文档：

如果两个单独的对象被给定为全局变量和局部变量，代码将被执行，就好像嵌入在类定义中一样。

因为我传入全局和 locals

  class Foo（object）：
 foo = [1,2,3 ] 
 @staticmethod 
 def bar（）：
 return foo [1] 
  

$ b

对于任何对解决方法感兴趣的用户，您可以注入命名空间回到命名空间['bar']。func_globals ¹ （受此启发）：

 code>>>>> namespace ['bar']。func_globals.update（namespace）
>>> namespace ['bar']（）
 2 
  

$ b

^{1 这将是 namespace ['bar'] .__ globals __。update on python3.x}

I'm a little curious why the following code raises a NameError.

>>> s = """
... foo = [1,2,3]
... def bar():
...    return foo[1]
... """
>>> namespace = {}
>>> exec(s, {'__builtins__': None}, namespace)
>>> print namespace
{'foo': [1, 2, 3], 'bar': <function bar at 0x7f79871bd0c8>}
>>> namespace['bar']()

At the normal interpreter level, we can find foo in bar.func_globals or bar.func_closure if in a function. I guess I'm wondering why namespace['bar'] doesn't put foo in func_closure ...

解决方案

It turns out that the answer was there all along in the docs:

If two separate objects are given as globals and locals, the code will be executed as if it were embedded in a class definition.

Since I'm passing in both globals and locals, it executes as if it were in a class.

class Foo(object):
    foo = [1,2,3]
    @staticmethod
    def bar():
       return foo[1]

not surprisingly doesn't work either :).

For anyone interested in a workaround, you can inject namespace back into namespace['bar'].func_globals¹ (inspired by this):

>>> namespace['bar'].func_globals.update(namespace)
>>> namespace['bar']()
2

Nice.

^{1It would be namespace['bar'].__globals__.update on python3.x}

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