# Learning to Code: Day 53 — Basic JavaScript Part 18

Hello all! A fine good evening to you tonight, let’s get back into this *JavaScript* business then shall we? Let’s finish up on a bit more *Recursion* for today, keeping it short and sweet. Cheers to* FreeCodeCamp* for the lessons.

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## Using Recursion to create a Countdown

If like me you’ve *always* wanted to create a countdown with *Recursion* through *JavaScript*, well you’ve come to the right place buddy!

We now know how to use recursion instead of a **for** loop, now let’s look at a function which by using recursion, returns a series of integers into an **array**, starting from 1.

Firstly we need a ** base case**: Where the return value is already known and the recursive function no longer needs to call itself.

Secondly, we need a ** recursive call**: This executes the original function with different arguments each time. To paraphrase

*FreeCodeCamp*here: if written correctly, “

*the base case is eventually reached*”.

So, let’s say you want to create a function which returns a series of integers into an **array**, starting from 1 until *n*, the function will need to accept an argument which states that number *n* is the last stop in the recursion. From there, it needs to recall itself in progressively smaller numbers of *n* until it reaches 1. Let’s take a look at that:

**Note**: The keyword “**const**” means “constant”, as in, what I write/define next will not be modified later on. This is different to **var** as **var** definitions can be modified and adjusted later on, which can cause bug problems in larger code.

Spot the obvious here? The **array** is *increasing*, however the recursive function is *decreasing*, this outcome is due to the fact that **.push()** happens *after* each recursive call has returned. At the moment that *n* is pushed into the** array**,** (n-1) **has already been evaluated and returned.. **[1,2,3,..n-1]**. So I think we can say that **.push()** is happening slower than the recursion is taking effect.

But what if we want to do this in reverse and have a *descending* **array**? Well, we can change a small part of this code example above, where the **array** was *ascending* and **.push()** put the numbers on the *end*, let’s use **.unshift()** (note that change on line 6 below) to “push” the numbers on the *beginning*, this way, we’ll have a *descending* **array**:

## Creating a Range with Recursion

If we have *two* perimeters in the **function** instead of *one* (*n* on line 1 above), how would this change the outlook of the function?

Let’s say we want to specify a range between point A and point B, instead of defining a predefined starting point of 1 up to *n*, like above. Let’s look at the code first then break it down:

So on line 1 we have our two perimeters - the starting point **startNum** and ending point **endNum**. Those are our two perimeters that were working inside of. If the user inputs two equal numbers, we want the **array** to output that single number (this is outlined in lines 2 and 3). Line 2 says “**If** the **starting number** *minus* the **ending number** is *strictly equal to* 0, then **return** the **second number**” (or the **starting number**, it wouldn’t matter, as they are the same in this case).

Next, from line 4 in the **else **statement, we say “**otherwise**, take a **variable **called **numbers** which takes the recursive call of (**startNum**, **endNum** - 1), and **push** that new **endNum** onto the *end* of the **array**”. This is effectively saying, take the higher number (**endNum**), decrease it by 1, and *add* that new **value** into the **array** sequence as each recursion occurs. The result of which is an *increasing* **array** from point A (**startNum**) to point B (**endNum**) and all the whole integers in-between.

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Aaaaaaaand let’s keep it short today. Thanks for checking in! See you next time for more on *JavaScript*.