Command exec in a function, name error - python3

Question

I'm begginer in Python and I'm in front of a problem that I can't understand. I tried to just define a variable with a exec() and then just print it. And it's working well. BUT when I do the same code in a function, it's not working...

Example :

def fonct():
    possibilite = [0,1,2,3]
    testc = 0
    testl = 1
    commande = "proba"+str(testc+1)+str(testl)+" = possibilite"
    exec(commande)
    print(commande)
    print(proba11)

The same thing but not in a function has for result, that the command print(proba11) returns [0,1,2,3] so it works. But for the example I got this :

proba11 = possibilite
NameError: name 'proba11' is not defined

There is no stories about globals or locals, everything is local...

Answer 1

Updating the local variables with exec() in Python 3 is tricky due to the way local variables are stored. It used to work in Python 2.7 and earlier.

To workaround this, you need to

• Pass an explicit locals dictionary to exec
• Grab the newly defined variable from the updated locals dictionary

Like so:

def fonct():
    possibilite = [0, 1, 2, 3]
    testc = 0
    testl = 1
    varname = "proba" + str(testc + 1) + str(testl)
    commande = varname + " = possibilite"
    _locals = locals()
    exec(commande, globals(), _locals)
    proba11 = _locals[varname]
    print(proba11)

Which works as expected.

You can read more about it here:

• exec() and variable scope
• How does exec work with locals?