1.4 KiB
1.4 KiB
id, title, challengeType, videoUrl, forumTopicId
id | title | challengeType | videoUrl | forumTopicId |
---|---|---|---|---|
56105e7b514f539506016a5e | Count Backwards With a For Loop | 1 | 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.
--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);
}