Python:为什么我不能使用 locals() 修改函数内的当前作用域？

Question

为什么创建/修改 locals() 的成员在函数内不起作用？

Python 2.5 (release25-maint, Jul 20 2008, 20:47:25)
[GCC 4.1.2 20061115 (prerelease) (Debian 4.1.1-21)] on linux2
Type "help", "copyright", "credits" or "license" for more information.

>>> # Here's an example of what I expect to be possible in a function:
>>> a = 1
>>> locals()["a"] = 2
>>> print a
2

>>> # ...and here's what actually happens:
>>> def foo():
...  b = 3
...  locals()["b"] = 4
...  print b
...
>>> foo()
3

Answer 1

为什么会这样？它旨在返回一个表示，并且从未打算用于编辑本地人。它永远不能保证作为一种工具工作，如 documentation警告。