exec() doesn't work with local variables in function [Python]

Question

I need exec() to modify a local variable. I read in somewhere and in documentation that including locals() would help it but it doesn't. So I'm kinda stumped! Check out my code to see what I mean.

https://code.sololearn.com/c4fAlxoJYEi3/?ref=app

Answer 1

Tashi N

def test3():
d=1
ldic=locals()
exec("d+=1",globals(),ldic)
d=ldic["d"]
print(d) # it works! returns 2

test3()
#ah thanks Tashi! The chosen awnser on stack overflow worked. however I needed to have the variable before hand unlike the example, but that's what I want anyway.

Answer 2

Yes, works for
global b
b=1

I'm quite sure there is a way to pass the scope to the exec function alternatively.
But I guess that's exactly the task: you want it local.

Phew, found it Ahri Fox
https://stackoverflow.com/questions/1463306/how-does-exec-work-with-locals

Answer 3

Ahri Fox Great, was about to post this because I saw it didn't work for you in first place, so I add this anyways ;)

https://code.sololearn.com/c9RnKkPL2F44/?ref=app

Answer 4

Wasn't it something with passing the scope in the method arguments?
globals or something?
Argh, can't remember. I'll search for it.

Answer 5

Pegasus check the code, locals don't work in functions
https://code.sololearn.com/c4fAlxoJYEi3/?ref=app

Answer 6

👍

Answer 7

Just played around with it

Answer 8

Zohir Yep, that output is proof.
First one is 2, which means it's working correctly, since exec() is adding 1 to variable a, which is 1.

Then comes the problem, I do the same code in a function, and it outputs 1, meaning it didn't add anything.

I try a recommended solution in another function, and it doesn't work, as evidenced by the 1.

Then I do Tashi's solution linked in a stack overflow link, and it works! it adds 1 which results in 2 inside a function.

So yeah, your original statement that it works for you is FALSE considering that output.

Answer 9

Tashi N In python 3.x its not possible because motivations writed in your posted link... Only in python 2.x... Anyway i think that passing an dict and reference it in exec code its sufficient for Ahri Fox needs

Answer 10

I thought exec() works for local variables as well?

Answer 11

Ace
def test3():
d=1
codeobj=compile("d+=1",'sumstring','exec')
exec(codeobj)
print(d)

test3()
#this returns 1. But I might be using compile() it wrong!

Answer 12

Dmitriy
no offense but, that's MY awnser. he copied my code from the question description

Answer 13

just pray around it and enjoy.

Answer 14

Joseph David Zamora Murillo
eval() wouldn't have helped in my situation (which isn't mentioned in this thread). I needed to execute code; not evaluate code. eval can't set variables. I needed to set/modify variables.

Answer 15

I like python but I need some complex questions

Answer 16

Dear Programmer,
exec() function works well in all conditions in Java for RunTime Environment.
But not sure about Python work.
Good Luck for best future.

Answer 17

compile() is the key!

Answer 18

Zohir this code is proof. try it on any machine. I've verified through code playground, another online REPL, and my own machine

https://code.sololearn.com/c4fAlxoJYEi3/?ref=app

Of course, the third test works correctly since it was the suggested awnser in this thread, which I added after the fact.

Answer 19

Ahri Fox ok

Answer 20

😊