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--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);
}