Javascript for loop behaviour in arrays

Question

I'm currently learning JS and doing some problems on Codewars. I've written a function to solve a problem, but I don't understand why the for loop in JS behaves the way it does in my solution, I'd be glad if anyone could help me. The problem is as follows:

Write an algorithm that takes an array and moves all of the zeros to the end, preserving the order of the other elements.

moveZeros([1, 2, 0, 1, 0, 1, 0, 3, 0, 1]) // returns[1, 2, 1, 1, 3, 1, 0, 0, 0, 0]

My solution is:

var moveZeros = function(arr) {
    let newarr = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] === 0) {
            newarr.push(arr[i])
        } else {
            newarr.unshift(arr[i])
        };
    }
    return newarr;
}

What my function returns is:

[1, 3, 1, 1, 2, 1, 0, 0, 0, 0]

Why does JS put 3 before 2, although its index in the array is 7?

Thanks in advance!

Answer 1

Well unshift() pushes the value in front of array.

Have a look at this example.ti understand how unshift() works.

let a = [];
 a.unshift(1);
 a.unshift(2);
 a.unshift(3);
 a.unshift(4);
console.log(a);

Push add values at the back of array.

look at example below to understand how push works.

let a = [];
     a.push(1);
     a.push(2);
     a.push(3);
     a.push(4);
console.log(a);

var moveZeros = function (arr) {
let newarr = [];
for (let i = 0; i < arr.length; i++) {
if (arr[i] === 0) {newarr.push(arr[i])}
else {newarr.unshift(arr[i])};
} return newarr;
}



let op = moveZeros([1, 3, 1, 1, 2, 1, 0, 0, 0, 0]);
console.log(op);