freeCodeCamp/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-javascript/count-backwards-with-a-for-loop.md

id, title, challengeType, videoUrl, forumTopicId, dashedName

id	title	challengeType	videoUrl	forumTopicId	dashedName
56105e7b514f539506016a5e	Count Backwards With a For Loop	1	https://scrimba.com/c/c2R6BHa	16808	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 count backwards by twos, 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.

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.

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

You should be using the array method push.

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

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

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

--seed--

--after-user-code--

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

--seed-contents--

// Setup
var myArray = [];

// Only change code below this line

--solutions--

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