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

id, title, challengeType, isHidden, videoUrl, forumTopicId

id	title	challengeType	isHidden	videoUrl	forumTopicId
56105e7b514f539506016a5e	Count Backwards With a For Loop	1	false	https://scrimba.com/c/c2R6BHa	16808

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.

Tests

tests:
  - text: You should be using a <code>for</code> loop for this.
    testString: assert(/for\s*\([^)]+?\)/.test(code));
  - text: You should be using the array method <code>push</code>.
    testString: assert(code.match(/myArray.push/));
  - text: <code>myArray</code> should equal <code>[9,7,5,3,1]</code>.
    testString: assert.deepEqual(myArray, [9,7,5,3,1]);

Challenge Seed

// Setup
var myArray = [];

// Only change code below this line

After Test

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

Solution

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