const "variable" undefined in following function

Question

Beginner here. Why userName is undefined in the function and how to make the function to have it as John Doe and not undefined?

const userName = 'John Doe';
console.log(userName);
const loggedInUser = (userName) => console.log('Logged in user is: ' + userName);
loggedInUser();

Console output:

index.js:22 John Doe
index.js:23 Logged in user is: undefined

You're hiding the "outer" `userName` with the `userName` parameter of the function. — Andreas, Feb 04 '18 at 09:19
[An example of variable shadowing in javascript](https://stackoverflow.com/questions/11901427/an-example-of-variable-shadowing-in-javascript) — Andreas, Feb 04 '18 at 09:26
FWIW, this is not a property, it's a "variable" though it would be more correct to say that it is a constant (since its value cannot change). — Felix Kling, Feb 05 '18 at 03:33

score 1 · Accepted Answer · answered Feb 04 '18 at 09:20

const loggedInUser = (userName) => console.log('Logged in user is: ' + userName);

defines a function named loggedInUser that takes a single argument.

After defining this function you call it on the next line loggedInUser();, but you don't provide the argument.

Try loggedInUser(userName);

1 Answers1