--description--
For loops don't have to iterate one at a time. By changing our final-expression
, we can count by even numbers.
We'll start at i = 0
and loop while i < 10
. We'll increment i
by 2 each loop with i += 2
.
const ourArray = [];
for (let i = 0; i < 10; i += 2) {
ourArray.push(i);
}
ourArray
will now contain [0, 2, 4, 6, 8]
. Let's change our initialization
so we can count by odd numbers.
--instructions--
Push the odd numbers from 1 through 9 to myArray
using a for
loop.
--hints--
You should be using a for
loop for this.
assert(/for\s*\([^)]+?\)/.test(__helpers.removeJSComments(code)));
myArray
should equal [1, 3, 5, 7, 9]
.
assert.deepEqual(myArray, [1, 3, 5, 7, 9]);
--seed--
--after-user-code--
if(typeof myArray !== "undefined"){(function(){return myArray;})();}
--seed-contents--
// Setup
const myArray = [];
// Only change code below this line
--solutions--
const myArray = [];
for (let i = 1; i < 10; i += 2) {
myArray.push(i);
}