Can't set a local variable in python directly through `locals()`

Question

I'll keep this one short and simple. Why does the following behaviour occur?

>>> globals()['x'] = 5
>>> x
5
>>> def f():
...    locals()['y'] = 7
...    y
...
>>> f()
Traceback (most recent call last):
  File "<pyshell#32>", line 1, in <module>
    f()
  File "<pyshell#31>", line 3, in f
    y
NameError: name 'y' is not defined

Here's an example of where this might be used:

import opcode
def foo():
    locals().update(opcode.opmap)
    #do stuff

I'm very curious why you want to do it this way instead of `y = 7`. — TheSoundDefense, Aug 04 '14 at 21:05
I don't think I would ever use this, I was just experimenting. — user3002473, Aug 04 '14 at 21:05
@TheSoundDefense I updated the question to include a scenario in which it might be useful. — user3002473, Aug 04 '14 at 21:10

score 3 · Accepted Answer · answered Aug 04 '14 at 21:05

3

The documentation for locals() includes:

Note: The contents of this dictionary should not be modified; changes may not affect the values of local and free variables used by the interpreter.

answered Aug 04 '14 at 21:05

Bill Lynch

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Well that was fast, thanks! – user3002473 Aug 04 '14 at 21:07
... and the reason you can't modify `locals()` is that the Python interpreter optimizes access to local variables at compile time; they aren't stored in a dictionary the way globals are. – kindall Aug 04 '14 at 22:17

Can't set a local variable in python directly through `locals()`

1 Answers1