Recursive question

I’m confused about a solution for a particular problem requiring recursion. I’ll post the problem and then cite my confusion below.

function flattenArray(arr) {
const newArr: string[] = [];
flatter(arr)
function flatter(arr){
arr.forEach((el)=>{
if(Array.isArray(el)){
    flatter(el)
} else{
    newArr.push(el)
}
})

}
return newArr
}

console.log(flattenArray([[["a"]], [["b"]]]));
console.log(flattenArray([1, [2], [3, [[4]]]]));

This algorithm works, but my confusion is over how. The function flatter is called before it is created. Within flatter, it is recursively called again in the if statement. This is twice.
Now, when I removed the first call of the flatter (the one before its instantiation in the next line) only empty brackets are returned.
I don’t quite I understand why the solution requires this first call of flatter AND how calling a local function before its instantiation is possible. Why a second call? Isn’t the call of flatter within the function enough?

Can someone please explain this to me?

Thank you.

I just thought of this…is this an example of closure?

this is a feature of how the function is defined. Defining the function in that way, with function funcName () {} you can call the function before declaring it.
see here: https://stackoverflow.com/questions/261599/why-can-i-use-a-function-before-its-defined-in-javascript


if you never call flatter, then newArr is never changed and you just return its starting value


if you never call the outer function then the function call inside that function never execute…

function func1 (where) {
   console.log("Function 1 called " + where);
}
function func2 (where) {
   func1("inside func2");
   console.log("Function 2 called " + where);
}

func1 is being called inside func2, but it you don’t call func2, will func1 ever be called? try yourself.
this is the same even if it is recursive. (if you want to try, be sure to give a stopping condition or there will be stack overflow)