---
id: 56105e7b514f539506016a5e
title: Count Backwards With a For Loop
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2R6BHa'
forumTopicId: 16808
dashedName: count-backwards-with-a-for-loop
---

# --description--

A for loop can also count backwards, so long as we can define the right conditions.

In order to decrement by two each iteration, we'll need to change our `initialization`, `condition`, and `final-expression`.

We'll start at `i = 10` and loop while `i > 0`. We'll decrement `i` by 2 each loop with `i -= 2`.

```js
var ourArray = [];
for (var i = 10; i > 0; i -= 2) {
  ourArray.push(i);
}
```

`ourArray` will now contain `[10,8,6,4,2]`. Let's change our `initialization` and `final-expression` so we can count backward by twos by odd numbers.

# --instructions--

Push the odd numbers from 9 through 1 to `myArray` using a `for` loop.

# --hints--

You should be using a `for` loop for this.

```js
assert(/for\s*\([^)]+?\)/.test(code));
```

You should be using the array method `push`.

```js
assert(code.match(/myArray.push/));
```

`myArray` should equal `[9,7,5,3,1]`.

```js
assert.deepEqual(myArray, [9, 7, 5, 3, 1]);
```

# --seed--

## --after-user-code--

```js
if(typeof myArray !== "undefined"){(function(){return myArray;})();}
```

## --seed-contents--

```js
// Setup
var myArray = [];

// Only change code below this line
```

# --solutions--

```js
var myArray = [];
for (var i = 9; i > 0; i -= 2) {
  myArray.push(i);
}
```