[New-bugs-announce] [issue4831] exec() behavior - revisited

David M. Beazley report at bugs.python.org
Sun Jan 4 15:52:37 CET 2009

New submission from David M. Beazley <beazley at users.sourceforge.net>:

Please forgive me, but I'm really trying to wrap my brain around the 
behavior of exec() in Python 3.   Here's a quote from the documentation:

   "In all cases, if the optional parts are omitted, the code is
    executed in the current scope."

This is referring to the optional use of the globals/locals parameters 
and seems to indicate that if they're omitted the code executes in the 
scope where the exec() appeared.

Yet, this code fails:

def foo():
    exec("a = 42")
    print(a)         # NameError: a

Now, I realize that exec() became a function in Python 3.  However, 
regardless of that, is it really the intent that exec() not be allowed 
to ever modify any local variable of a function?   In other words, do I 
really have to do this?

def foo():
    ldict = locals()
    a = ldict['a']

I submitted a bug report about this once before and it was immediately 

I would appreciate some greater clarity on this matter this go around.  
Specifically, what is the approved way to have exec() modify the local 
environment of a function?

components: Interpreter Core
messages: 79059
nosy: beazley
severity: normal
status: open
title: exec() behavior - revisited
type: behavior
versions: Python 3.0

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