Python：为什么不能使用locals（）修改函数中的当前作用域？

Question

为什么在函数中创建/修改locals（）的成员不起作用？在

Python 2.5 (release25-maint, Jul 20 2008, 20:47:25)
[GCC 4.1.2 20061115 (prerelease) (Debian 4.1.1-21)] on linux2
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>>> # 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

locals（）返回名称空间的副本（这与globals（）的操作相反）。这意味着对locals（）返回的字典执行的任何更改都将无效。在示例4.12签入dive into python。在