Any way to modify locals dictionary?

by

I just tested exec and it works in Python 2.6.2

>>> def test():
...     exec "a = 5"
...     print a
...
>>> test()
5

If you are using Python 3.x, it does not work anymore because locals are optimized as an array at runtime, instead of using a dictionary.

When Python detects the “exec statement”, it will force Python to switch local storage from array to dictionary. However since “exec” is a function in Python 3.x, the compiler cannot make this distinction since the user could have done something like “exec = 123”.

http://bugs.python.org/issue4831

To modify the locals of a function on
the fly is not possible without
several consequences: normally,
function locals are not stored in a
dictionary, but an array, whose
indices are determined at compile time
from the known locales. This collides
at least with new locals added by
exec. The old exec statement
circumvented this, because the
compiler knew that if an exec without
globals/locals args occurred in a
function, that namespace would be
“unoptimized”, i.e. not using the
locals array. Since exec() is now a
normal function, the compiler does not
know what “exec” may be bound to, and
therefore can not treat is specially.

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