I am doing an if statement with multiples of a number

Question

I am doing an exercise, the problem is that my if/else structure does not work properly and I do not know why.

Here is the exercise statement and my code

Your task is to write a function, fizzBuzz, that accepts a number and returns a string:

'fizz' if the number is divisible by 3;
'buzz' if the number is divisible by 5;
'fizzbuzz' if the number is divisible by both 3 and 5.
'{number}' if the number doesn't fulfil any of the above conditions.

function fizzBuzz(number) {

  if (number % 3 === 0) {
    return "fizz"
  };
  if (number % 5 === 0) {
    return "buzz"
  };
  if (number % 3 === 0 && number % 5 === 0) {
    return "fizzbuz"
  };
  else return number
}

This is a good opportunity for you to start familiarizing yourself with [using a debugger](https://stackoverflow.com/q/25385173/328193). When you step through the code in a debugger, which operation first produces an unexpected result? What were the values used in that operation? What was the result? What result was expected? Why? To learn more about this community and how we can help you, please start with the [tour] and read [ask] and its linked resources. — David, Oct 13 '22 at 17:17
Hint: Under what condition would the third `if` statement ever be reached at all? — David, Oct 13 '22 at 17:17

score 0 · Answer 1 · answered Oct 13 '22 at 17:44

Try a little mental debugging. Look at your code and run different values through it in your mind:

What happens if you run the value 6 through it?
What happens if you run the value 10 through it?
What happens if you run the value 15 through it?

Ask yourself, "How could I fix this?" (hint: order of operations is important).

Something like this would do what you expect:

function fizzBuzz(n) {
  const fizz = n % 3 ? '' : 'fizz' ;
  const buzz = n % 5 ? '' : 'buzz' ;

  return !fizz && !buzz ? '{number}' : `${fizz}${buzz}` ;

}

or this:

function fizzBuzz( n ) {
  let s;

  if ( n % 3 === 0 ) {
    s = 'fizz' ;
    if ( n % 5 === 0 ) {
      s = 'fizzbuzz' ;
    }
  } else if ( n % 5 == 0 ) {
    s = 'buzz' ;
  } else {
    s  = '{number}' ;
  }

  return s;
}

This does [at most] 2 divisions and 2 comparisions.

The former does 2 divisions and 3-4 comparisons, so it is nominally less efficient.

But I know which version I'd rather look at.

zer00ne · Answer 2 · 2022-10-13T17:59:57.883

The problem is, if a number is divisible by 5 and 3 (ex 15) "fizz" would only be returned (as a factor of 3) because every block {...} terminates the function with a return (this technique is called a "short-circuit", which isn't a bad practice). So you need to put the 5 and 3 condition first. Also that condition had an undefined variable called solution so that would've been the first error. One minor thing is that each block {...} was suffixed with a semi-colon: ; which is just bad formatting but it doesn't affect functionality.

function fB(number) {
  if (number % 3 === 0 && number % 5 === 0) {
    return "fizzbizz";
  }
  if (number % 3 === 0) {
    return "fizz";
  }
  if (number % 5 === 0) {
    return "bizz";
  }
  return number;
}

console.log(fB(15));
console.log(fB(575));
console.log(fB(49));
console.log(fB(51));
console.log(fB(80));
console.log(fB(375));
console.log(fB(99));
console.log(fB(Infinity));

I am doing an if statement with multiples of a number

2 Answers2