# Recursion - questions/help/guidance

Tell us what’s happening:

So I was able to do the solution just from mirroring the example provided… however I have no idea whats going on. I was hoping someone could help explain what js is doing in simpler terms. I’m trying to grasp recursion at the moment and my head hurts thinking about it.

so, I pass the arguments array and 3, it checks if 3 == 0 which it doesn’t, so then moves to the else statement. It takes in array and n - 1 (2?), then adds arr[n-1] so the second index of the array (4?) so 6 so far? And then it checks the first condition, n is at 2 now so moves on to the else. n-1 (1) + arr[1] which is 3…and i guess that does equal 9. so now n is at 1, it goes to else statement n-1 = 0, adds that to index 0 which is 2… so shouldn’t it end at 11? Or does it see the n -1 = 0 and go back to the if statement to end it?

I’m also trying to comprehend where this is stored in memory. Since the total adding up isn’t assigned to any variable. does js just know to hold that value and keep adding to it?

I’m struggling with the self teaching at the moment I also didn’t get much sleep last night and doing this alongside my real job so apologies if i missed something simple!

``````
let array = [2, 3, 4]

function sum(arr, n) {

if (n == 0) {
return 0
} else {
return sum(arr, n - 1) + arr[n - 1]
}

}

console.log(sum(array, 3))
``````

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Challenge: Replace Loops using Recursion

Hi Eric.
When you’re using recursion each function call is maintained on the stack until it finally returns. Which means that its state is maintained in memory until it finally finishes off.

As for the value that it ends up with its because each run is passing the returnValue + arr[n - 1] along with it.

We’re returning the value of whatever sum returns + arr[n-1].
Until we reach the base case of n == 0 we never actually have a value for sum().

All of this looks like this. // Fixed
? Represents the unknown value because we haven’t returned anything out of any of the layers of sum yet. This is why we keep calling the function sum until n = 0.
sum -> ? + arr[n - 1] // n - 1 = 2
sum -> sum -> ? + arr[n - 1] // n - 1 = 1
sum -> sum -> sum -> ? + arr[n - 1] // n - 1 = 0
sum -> sum -> sum -> sum -> basecase = return 0
sum -> sum -> sum -> return (0 + arr[0]) == 0 + 2
sum -> sum -> = return (2 + arr[1]) == 2 + 3
sum -> = return (5 + arr[2]) == 5 + 4
return == 9

I made some mistakes on how the recursion works and how the base case works.
I also suggest this computerphile video on recursion. https://www.youtube.com/watch?v=Mv9NEXX1VHc

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Thank you so MUCH! I am watching that video right now, this was extremely helpful!

You’re welcome
I did realize I had some mistakes with my answer so I will be editing my post to fix it.

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