Recursives are confusing. I know what code to write to pass but I do not understand the function

[ayindejam2009]

At what point is the array created? Shouldn’t a variable for the array be initialized first i.e. const countArray = ? Also if n decreases by 1 each time, where is that new number stored? Clearly it is pushing new values of n as it decreases by 1 but where in the function is it recursive? It seems to be doing some sort of n = n-1 thing to get the rest of the values in the array somewhere in the function but I cannot see an expression for this anywhere? It also says in the code description:

At first, this seems counterintuitive since the value of n decreases , but the values in the final array are increasing . This happens because the push happens last, after the recursive call has returned. At the point where n is pushed into the array, countup(n - 1) has already been evaluated and returned [1, 2, ..., n - 1] .

Ok, again, where are these new vaues of n stored so they are recalled?

  **Your code so far**

// Only change code below this line
function countdown(n){
if (n < 1) {
  return [];
} else {
  const countArray = countdown(n - 1);
  countArray.unshift(n);
  return countArray;
}
}

console.log(countdown(5));

// Only change code above this line

  **Your browser information:**

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.67 Safari/537.36

Challenge: Use Recursion to Create a Countdown

Link to the challenge:

freeCodeCamp.org

[JeremyLT]

@ayindejam2009

const countArray = countdown(n - 1);

This line stores the ‘smaller’ results.

Recursion is confusing, so struggling with it is normal.

[kinome79]

So:

const countArray = countdown(n-1);

does your recursion and creates your array… It keeps calling itself over and over, with a value of n-1. So if your first call was countdown(5), it will then call itself again as countdown(4), etc. As it keeps decreasing the value of n passed to the function with every call, eventually you get to n = 0… when n = 0, that function will return your blank array (your array is now created)… and then as the functions close out, each one adds the value of N to the front with unshift… meaning your array will build up as [1] then [2,1], then [3,2,1] until it gets back to your original function call, and your array is built. Hope that helps it make sense.