NameError: 'c' is not defined

Question

I write below code in python(version 3.6.2):

def test():
    temp = 'c=1'
    exec(temp)
    print(c)

test()

when run it, I get below error:

NameError: name 'c' is not defined

but when run below code :

temp='a=1'
exec(temp)
print(a)

it print 1.

How do I fix this?

You don't have a variable called c how you would print it?, print(c) raise the error. — Fady Saad, Aug 22 '17 at 18:26
I am unable to reproduce your error in python2.7 or python3. It prints `1` each time. — o-90, Aug 22 '17 at 18:27
The documentation on Python 3's [`exec()`](https://docs.python.org/3/library/functions.html#exec) is pretty clear: *"modifications to the default locals dictionary should not be attempted. Pass an explicit locals dictionary if you need to see effects of the code on locals after function exec() returns."* (globals is a different story) — Lukas Graf, Aug 22 '17 at 18:33
Also, you *almost certainly* don't want to be using `exec` to dynamically assign variables. There is a better way. — juanpa.arrivillaga, Aug 22 '17 at 18:34

score 0 · Answer 1 · answered Aug 22 '17 at 18:40

0

The exec() function can only be followed by a string or an object. What you asked it to do is not one of those. You should try just defining c in another way or stop using exec()

answered Aug 22 '17 at 18:40

Huh? OP passes a string, the problem isn't that, it's that `exec` does not modify `locals` – juanpa.arrivillaga Aug 22 '17 at 18:42

NameError: 'c' is not defined

1 Answers1