# Solution to Day 2 of 7 Days of JS

## 30 November 2019

Let’s see. First of all a solution using a while loop:

function sumboth(array) {
let count = array.length;
let i = 0, even = 0, odd = 0;
while (i < count) {
array[i] % 2 == 0 ? even += array[i] : odd += array[i];
i++;
}
return {even, odd};
}

console.log(sumboth([3,1,6,2]))


And as a second choice, a solution using a for loop:

function sumboth2(array) {
let count = array.length;
let even = 0, odd = 0;
for (let i = 0; i < count; i++) {
array[i] % 2 == 0 ? even += array[i] : odd += array[i];
}
return {even, odd};
}

console.log(sumboth2([3,1,4,2]))


I’ve used the conditional (ternary) operator:

condition ? inCaseIsTrue : inCaseIsFalse;


You can read more about it here.