I’m trying to figure out why we assign rangeOfNumbers to nums. I console.log’d nums and it returns undefined until numbers start to be pushed onto it. It seems like the nums is only affected by the following line:

First, the easiest thing to answer: You return an empty array if (endNum < startNum) because, if that ever happens, the function is getting inputs that it can’t handle. So if you run rangeOfNumbers(2, 1), it’ll return an empty array because the first number has to be less than the second number.

As for what’s happening in the else block, that’s the recursion bit… you sort of keep looping through the function until startNum === endNum, then it returns the nums array.

Yeah, recursion is one of those things that’s both hard to understand and hard to explain. And I’m not particularly good with the concept myself.

But let me try explaining it a little bit better:

So lets say we pass the numbers 2 and 3 to the function like rangeOfNumbers(2, 3) (this is function call #1):

First, it will hit that else block. And the first line of the else block is const nums = rangeOfNumbers(startNum, endNum - 1); The program needs to resolve this before it can move forward to the nums.push(endNum); line.

So now we’re now resolving the function rangeOfNumbers(2, 3 - 1) (this is function call #2). Because 2 < 2 is false, we go to the else block again. So now we have to resolve rangeOfNumbers(2, 2 - 1) (this is function call #3) before we can move forward.

Finally, 2 < 1 is true, so we have an empty array [] as the return value from function call #3.

So in function call #2, num = [], then we can move to the nums.push(endNum); line. In function call #2endNum = 2, so we push that onto num and we end up with [2] as the return value from function call #2.

So then function call #1 gets [2] as the value of num. In function call #1, endNum = 3. So we then push that onto the the num const.

So we end up with [2,3] as the final value of num which then gets returned from the original function call.

Ya, I get all that. I get what’s happening, for the most part. I’m just not exactly sure why we return an empty array or what happens with this array, if anything. It doesn’t seem to go to nums. I console.log’d nums and it never seemed to be an empty array at any point. It comes back as undefined until we start pushing numbers to it.

I also don’t get why we assign the function to nums . nums doesn’t seem to have any value until we push each number to it.

Thanks for taking the time to write that all out though.

I dunno. That’s what I’m trying to figure out. I’m starting to think that that the empty array IS being returned nums. How that works, I’m not sure. Intuitively, I would think it would return nums = ; or something like that. I’ll just assume that’s what’s happening.

I’m still not sure what assigning the function to nums does. Now that I think about it, that’s probably why it knows to return the empty array to nums. Maybe that’s all it is, if that is the case. I dunno.

I’m starting to think that the empty array IS being returned to nums, and that maybe it knows to return it to nums because we assign the function to nums. So maybe that answers everything, or at least part of it. I’m still not positive though.

It’s not assigning the function to nums. Its assigning the value returned by the function to nums. And the function always returns an array.

If you invoke the function with:

rangeOfNumbers(7,7);

Remember here that endNum equals 7.
You’ll hit the else and the following will execute:

const nums = rangeOfNumbers(7, 6);

Before it can go any further than this line it must return a value for rangeOfNumbers(7, 6). You know what that is because that will trigger the base case. So nums will be set to [].

Now we can go to the next line, which we can sub in values to make it:

[].push(7);

Which is [7] and that’s what the original function call returns.