* fix: Simplify wording to make more sense I think that "count odd numbers backward by twos" instead of "count backward by twos by odd numbers" is easier to understand. * Update curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-javascript/count-backwards-with-a-for-loop.md I think that this variation sounds better than what I originally wrote. Co-authored-by: Randell Dawson <5313213+RandellDawson@users.noreply.github.com> Co-authored-by: Randell Dawson <5313213+RandellDawson@users.noreply.github.com>
1.5 KiB
1.5 KiB
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 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.
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 backwards by twos to create an array of descending 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);
}