Why isn't the except NameError picking up the name error on my code?

Question

I'm trying to have a user input an integer, while trying to catch errors if they put in floaters or strings. Any help would be awesome guys!!

try:
    user_input = int(input())
except ValueError:
    print("Please enter a whole number with no decimal points")
except NameError:
    print("user_input not defined")

Does this answer your question? [Python NameError: name is not defined](https://stackoverflow.com/questions/14804084/python-nameerror-name-is-not-defined) — Andreas is moving to Codidact, Jul 26 '20 at 00:41
@JimmyCarlos As I wrote, in python 3 there is no way for `NameError` to be raised by this code (assuming `int`, `input` or `print` were not removed from scope, of course) — DeepSpace, Jul 26 '20 at 00:41
@Phoenix So you are using other code which you didn't show us. There is no way for the code you provided to raise `NameError` in Python 3 — DeepSpace, Jul 26 '20 at 00:43

score -1 · Accepted Answer · answered Jul 26 '20 at 00:37

-1

NameError is called when you try to access a variable not declared yet. But in this code, you will always create user_input, so the try block can't go to the nameerror block.

answered Jul 26 '20 at 00:37

JimmyCarlos

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If OP is using Python 2 then `NameError` is raised (and caught, so the question is still not clear either way) if a non-existing variable name is entered when `input` is called. – DeepSpace Jul 26 '20 at 00:40

Why isn't the except NameError picking up the name error on my code?

1 Answers1