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/** | |
Code Challenge Needed: | |
Using Javascript, given an array of n integers (example: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 112, 113] ), remove all odd numbers, leaving only the even numbers. | |
Rules: | |
NO LOOPING. This means native methods, or libraries that loop for you are not allowed either. | |
Supply the answer on github via Gist | |
**/ | |
var numbersArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 112, 113]; | |
//initial array | |
console.log('The original numbers array was : ', numbersArray); | |
//this code is just one more variation( looping with a for... I know :) ) but despite the polemic about performace this kind of solutions are more clear, more maintainable and reusable code also are more easy to trace | |
function getEvens(array){ | |
let position = 0, | |
arrayLength = array.length; | |
for (var i = 0; i < arrayLength; i++) { | |
if ( !(array[i] % 2) ) { | |
array[position++] = array[i]; | |
} | |
} | |
array.length = position; | |
return array; | |
} | |
var evensArray = getEvens(numbersArray); | |
//final array | |
console.log('The evens numbers are : ', evensArray); |
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