Last value of my FizzBuzz function returns undefined

Question

I'm trying to run the FizzBuzz problem solving, it seems to be working, however when it reaches the user input number it returns undefined.

Here is my code:

const prompt = require("prompt-sync")();

let userChoice = parseInt(prompt("Please enter a number: "));

function fizzBuzz () {
    for (let i=1 ; i <= userChoice; i++){
        if (i % 3 === 0 && i % 5 ===0) {
            console.log("FizzBuzz");
        }
        else if (i % 3 === 0){
            console.log("Fizz"); 
        }
        else if (i % 5 === 0){
            console.log("Buzz");
        }
        else {
            console.log(i); 
        }
    }
}

console.log(fizzBuzz(userChoice));

This is the outcome in the console :

Remove the `console.log` from the last line. `fizzBuzz` is taking care of logging things along the way. It doesn't return anything to be output. Just call `fizzzBuzz(userChoice);` — StriplingWarrior, Aug 16 '21 at 19:20
You're trying to pass a parameter into the fizzBuzz function (`fizzBuzz(userChoice)`) but your function doesn't accept a parameter (`function fizzBuzz ()`) — James, Aug 16 '21 at 19:23
@James `let` is block-scoped, so `userChoice` is still accessible within the function, but yeah, good idea to add it to the param list for `fizzBuzz()`. — BenM, Aug 16 '21 at 19:29
Hi all, thanks for the help, So I removed the console.log and called the function at the end, it seems to be working now, AdditionalI tried Ben first code option and it wont run at all, sadly — Diana C, Aug 16 '21 at 19:35

BenM · Answer 1 · 2021-08-16T19:27:13.657

You need to actually return a value from your fizzBuzz() function. At the moment, that function is returning nothing, so when you log its return value, you're getting undefined.

function fizzBuzz (userChoice) {
    let output = [ ];

    for (let i=1 ; i <= userChoice; i++){
        if (i % 3 === 0 && i % 5 ===0) {
            output.push("FizzBuzz");
        }
        else if (i % 3 === 0){
            output.push("Fizz"); 
        }
        else if (i % 5 === 0){
            output.push("Buzz");
        }
        else {
            output.push(i);
        }
    }

    return output.join("\n");
}

Alternatively, just don't wrap your actual function call in a console.log, so your last line will simply be:

fizzBuzz(userChoice);

Last value of my FizzBuzz function returns undefined

1 Answers1