Why running exec inside function fail in python3?

Question

Results running the code for python3:

------- input option 1 - exec code is executed --------
0 - for running inside function, 1 - for running in main programa: 1
option = 1
10

------- input option 0 - exec code not executed inside function ----
0 - for running inside function, 1 - for running in main programa: 0
option = 0
code inside execfunction => A = 10
Traceback (most recent call last):
  File "myexec.py", line 19, in <module>
    print(A)    
NameError: name 'A' is not defined
---------------------Code --------------------------

myexec.py

def execfunction(icode):
    print('code inside execfunction => %s' %(icode))
    exec(icode)

option = int(input("0 - for running inside function, 1 - for running in main programa: "))

print('option = %d' %(option))

code = 'A = 10'

if (option == 1):
    exec(code)
else:
    execfunction(code)  

print(A)

The code does not "fail". The variable has a different scope if the code is executed within the function. Move the print inside the function and you will see. — Mike Scotty, May 31 '18 at 13:23
@MikeScotty Thats what I thought, that `A` was a local variable of the function `execfunction()` but I tried to `print(A)` and it still doesn't work, very weird. idk why. — Flaming_Dorito, May 31 '18 at 13:27
#myexec.py def execfunction(icode): print('code inside execfunction => %s' %(icode)) exec(icode) print(A) option = int(input("0 - for running inside function, 1 - for running in main programa: ")) print('option = %d' %(option)) code = 'A = 10' if (option == 1): exec(code) print(A) else: execfunction(code) — Marco Aurélio Falcão, May 31 '18 at 13:30
Sorry, it seems I made an assumption w/o actually testing. Even though ``A`` is listed in ``locals()`` inside ``execfunction``, you cannot simply print it. @DavidZarebski is right with linking to [https://stackoverflow.com/a/15087355/4349415](https://stackoverflow.com/a/15087355/4349415) — Mike Scotty, May 31 '18 at 13:36

score 0 · Answer 1 · answered May 31 '18 at 13:30

0

Is it because exec is a function in python 3 but a statement in python 2? Have a look at this discussion

answered May 31 '18 at 13:30

zar3bski

2,773
7
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No, that wouldn't matter in this case. That only matters when passing exec itself as a function. E.g., `map(exec, ["a = 1", "b+=2", "c=3"])` would work in Python 3, but fails in Python 2. His issue is with scoping. – Edward Minnix May 31 '18 at 13:42
Thanks for your attention! I have read the discussion. The workaround suggested is too complicated if one wants to keep his code easy to understand. – Marco Aurélio Falcão May 31 '18 at 13:55
Edward, Thanks, right, scoping. Maybe the compatibility with python3 should have been kept in this case. – Marco Aurélio Falcão May 31 '18 at 14:00

score 0 · Answer 2 · answered May 31 '18 at 13:37

0

Python compiler will LOAD GLOBAL on encountering the print statement, where it fails as undefined variable 'A'. If you try to disassemble your code [import dis], you will get to see the execution of the backend process call.

A nice and well defined explanation is given in this Creating dynamically named variables in a function in python 3 / Understanding exec / eval / locals in python 3

answered May 31 '18 at 13:37

Nishant Patel

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thanks for your link! I have read the link and it seems to be too complicated for a thing that should be simple – Marco Aurélio Falcão May 31 '18 at 13:48

score 0 · Accepted Answer · answered May 31 '18 at 13:48

If you really want execfunction to execute the function in the global scope, you can do

def execfunction(code):
    exec(code, globals())

Then, this will make the calls to execfunction execute in the global scope, instead of only in the local scope of the function.

For reference: https://docs.python.org/3/library/functions.html#exec

Why running exec inside function fail in python3?

3 Answers3