Javascript array reverse confusion

Question

In the following example, when I log to the console, I get ["c","b","a"] for all three arrays.

let x = Array.from("abc");
let y = x;
let z = y.reverse();
console.log({x:x, y:y, z:z})

y = ["c", "b", "a"] because Array.prototype.reverse() reverses the array in place

z = ["c", "b", "a"] because I reversed y and then assigned it to z

but why does x = ["c", "b", "a"]?

you do not assign a new array, but took the same object reference. — Nina Scholz
– Nina Scholz, Commented Oct 20, 2020 at 18:24

CertainPerformance · Accepted Answer · 2020-10-20 18:24:36Z

1

The script only ever creates a single array:

let x = Array.from("abc");

The Array.from creates the array and assigns it to x.

let y = x;

The above creates the y variable name and points it to the same array.

let z = y.reverse();

The above reverses the array in-place, and also creates the z variable name and points it to said array.

The array is never cloned, so x, y, and z are all referring to the same array in memory.

answered Oct 20, 2020 at 18:24

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2 Comments

I figured this is what was happening. Is there a way to check whether two variables point to the same object in memory?

Something like x === y

phi-rakib · Accepted Answer · 2020-10-20 18:24:11Z

1

when you are assigning y = x you are passing reference of x. So copy x by spread operator.

let x = Array.from("abc");
let y = [...x];
let z = y.reverse();
console.log({x:x, y:y, z:z})

answered Oct 20, 2020 at 18:24

phi-rakib

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