Replace Loops using Recursion -- hung up

function sum(arr, n) {

if(n <= 0) {

return 0;

} else {

return sum(arr, n - 1) + arr[n - 1];

}

}

console.log(sum([2, 3, 4, 5], 3))

I don’t understand why "sum(arr, n-1) " is not being added to the total. It seems that the function is just looping through and adding arr[2], arr[1] and arr[0], while "sum(arr, n-1) " is merely functioning as a counter.

How is it only functioning as a counter when it’s on the left side of an equation…that is, sum(arr, n-1) + arr[n-1] . The only thing that is being added is what is being looped on the right side of the equation.

I suspect that part of what makes this confusing is that you are used to thinking in terms of loops. This function doesn’t use a loop or counters.
This line:

return sum(arr, n - 1) + arr[n - 1];

is adding two numbers together: the number in the array (arr[n-1]) and the result of calling the function sum(arr, n-1).
sum(arr, n - 1) will resolve to a number before it is added to arr[n-1]. You’ve seen this sort of thing before when you’ve used built in methods - Math.random() + 1 is an example where the function Math.random() returns a number and that is what is added to 1.

Recursive functions call themselves, but it’s important to remember that a function isn’t a single entity like an assembly line - it’s a set of instructions. When sum calls itself it isn’t just going back to the top (like a loop does). It is creating a completely separate function call.

sum(arr, n - 1) + arr[n - 1] calls a function, waits for it to return, and then adds a number to it. sum(arr, n - 1) executes and returns a number. That number might be zero or it might not. If the second argument it was passed wasn’t zero, then it will call also have to call the sum function and wait for it to finish before it returns a number. It doesn’t know or care whether it was called recursively.

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Thank you for taking the time to respond, it makes more sense now :slight_smile:

I’m glad I could help. Happy coding!

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