I'm trying to solve FizzBuzz with a recursive, pure-function-style approach. My local test produces the expected output, but the online judge reports `TypeError: test is not a function` at `arr[i - 1] = test(i)`. The callback is declared and passed into `fizzBuzz`, so I'm not sure why the judge cannot call it. Here is the relevant code:
```js
const test = (i) => {
if (i % 5 === 0 && i % 3 === 0) return "FizzBuzz";
if (i % 5 === 0) return "Buzz";
if (i % 3 === 0) return "Fizz";
if (i % 3) return i + "";
};
const fizzBuzz = (i, n, test, arr) => {
arr[i - 1] = test(i);
if (i !== n) fizzBuzz(i + 1, n, test, arr);
return arr;
};
```
It works when I call it myself with all four arguments, but fails when submitted to the judge.
2 Answers
The issue is probably the required function signature, not the callback itself. The judge calls your submitted function with only one argument, `n`, because that is the expected FizzBuzz interface. In your version, that value becomes `i`, while `n`, `test`, and `arr` are `undefined`. Consequently, `test(i)` tries to call `undefined` as a function. Keep helper functions inside the submitted function or call your recursive function with all of its required arguments from a wrapper matching the platform’s expected signature.
Try implementing the required `fizzBuzz(n)` entry point and create the array and callback inside it. For example:
```js
var fizzBuzz = function(n) {
const result = [];
for (let i = 1; i <= n; i++) {
if (i % 15 === 0) result.push("FizzBuzz");
else if (i % 3 === 0) result.push("Fizz");
else if (i % 5 === 0) result.push("Buzz");
else result.push(String(i));
}
return result;
};
```
Also make sure the non-multiple case always returns a string. Your original `if (i % 3)` happens to work for numbers that are not divisible by three, but an unconditional final `return String(i)` is clearer and safer.
The parameter name `test` shadows the outer constant with the same name, but JavaScript permits that and it works when all four arguments are supplied. Renaming it to something like `formatter` would make the code clearer, but it does not explain the judge error by itself.