Array reversing function JS

Question

A beginner JS question.. I need to write a function that reverses an array that goes as a function's input. (I cannot use a reverse method).

I wonder why this works:

function reverseArrayInPlace(array) {
 for (let i = 0; i < Math.floor(array.length / 2); i++) {
 let old = array[i];
 array[i] = array[array.length - 1 - i];
 array[array.length - 1 - i] = old;
 }
 return array;
}
let arr = [0, 1, 2, 3, 4, 5];
console.log(reverseArrayInPlace(arr))

But this does NOT:

function reverseArrayInPlace(arr) {
 let len = arr.length;
 for (counter = 0; counter < 2 * len; counter += 2) {
 arr.unshift(arr[counter]);
 }
 arr = arr.slice(0, len);
}
let b = [0, 1, 2, 3, 4, 5];
console.log(reverseArrayInPlace(b));

Looks like arr = arr.slice(0,len); part is not working..I wonder why when:

b = b.slice(0,6);
[5, 4, 3, 2, 1, 0]

• Hanlet Escaño: Set a breakpoint and look at what's happening to the values in the array.
• sbattou: console.log() and breakpoints are your friends
• VLAZ: You are not returning anything from the function. If you return arr at the end, you'll see it works.
• Heretic Monkey: Possible duplicate of javascript reverse an array without using reverse()
• quirimmo: @LeviBerkowitz there is not a "right way" . What does "right way" mean?

Answer 1

If you want to change the input arrray, avoiding return, use splice:

function reverseArrayInPlace(arr) {
 let len = arr.length;
 for (counter = 0; counter < 2 * len; counter += 2) {
 arr.unshift(arr[counter]);
 }
 arr.splice(len);
}
var b = [0, 1, 2, 3, 4, 5];
reverseArrayInPlace(b);
console.log(b);

EDIT:

If you want to do something like:

console.log(reverseArrayInPlace(b));

your function MUST return something, otherwise the print will always be undefined, even if b has been reverted

• Jonas Wilms: This works by accident as arr is dropped. Remove arr = and it will still work.

Answer 2

 arr = arr.slice(0, len);

slice returns a new array which you store inside the local variable arr, that does not change b, which is still pointing to the whole array. To mutate the array you could:

 arr.splice(len, len);

and then you have to return arr to log something meaningful.

• Jonas Wilms: @quirimmo nope, variables are passed by value.
• VLAZ: Well, arrays are objects passed by value...where the value is their reference. So, doing in-place modification of arr will also change it outside the function but arr = /* something */ would not affect the array outside the function.
• quirimmo: yeah that's what I meant, sorry bad reading. If he wants to change the input arr, avoiding returns, with splice he can directly modify it
• quirimmo: then of course if he wants to do console.log(reverseArrayInPlace(b)); the method must return it, otherwise the value printed will always be undefined because the function does not return anything

Answer 3

Because the array is passed to the function by copying the reference, hence you cannot change the external reference from inside the function.

• Félix Adriyel Gagnon-Grenier: Not sure why that was downvoted. This is accurate, and explains the reason.