Replace Loops using Recursion - Base case

Hello! I was going through Basic JavaScript - Replace Loops using Recursion, and couldn’t understand the base case of the example (multiply function).

function multiply(arr, n) {
    if (n <= 0) {
      return 1;
    } else {
      return multiply(arr, n - 1) * arr[n - 1];

I get how and why it works when n > 0. But if I call this function with n as 0,

multiply(arr, 0);

It will return 1. Isn’t it simply wrong? Shouldn’t it return a string of error message or something?

Or is it possible to do this

if (n <=0) {
   return "Wrong input";

and set base case differently?
Thanks in advance!

You want it to return 1 because the function calls itself with (n - 1). If I call the function with 1, then the function multiplies the array by 1, moves to 0, and returns 1.

So, let’s do multiply([1, 2], 2).
First it returns multiply([1, 2], 2 -1) * arr [2-1], or multiply([1, 2], 1) * arr[1].
Now it’s calling the function again. We look at that, and we get:
multiply([1, 2], 0) * arr [1-1] * arr[2-1)

Finally, it calls the function with 0. Now, if we use your suggested base case of returning “wrong input”, we get:
"wrong input" * arr[0] * arr[1], which evaluates to NaN.

If we use the suggested base case of return 1, then we get 1 * arr[0] * arr[1], which yields the expected result of 2.

Does this make sense?

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you would need to separate the recursion base case, where you need the 1, or arr[0] as starting value, and the invalid input thing, otherwise you would find yourself with "Wrong input" * 5 (or any other number) and that results in NaN when the recursion is using a function call with a lower n to calculate the current call


Thanks for the detailed explanation @nhcarrigan, @ieahleen, I will try writing different base case!

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