How come if a = b, then b = a?

Question

Let me explain. I have the following piece of code :

let a = [1, 2, 3];
let b = a;
b.splice(2, 1);
console.log("a: " + a);
console.log("b: " + b);

I would have expected to get something like :

a = [1, 2, 3];
b = [1, 2];

However, after running the code, it turns out both "a" and "b" equal [1, 2]. I am really confused, since "b" should only have been assigned a copy of "a", instead of acting as some sort of a pointer to "a". I wonder if it is because JS handles arrays (objects) differently of if it is something specific to the splice function. I would also like to know how you would bypass that odd behavior.

Thanks for your time.

`b` references `a`, it is not a copy. Someone will find the dupe. — epascarello, Jan 02 '19 at 15:14
*"since "b" should only have been assigned a copy of "a"* Why ? Where are you copying ? — Denys Séguret, Jan 02 '19 at 15:14
You say *"b should only have been assigned a copy of a"* but your code shows b being assigned a, not a copy. — Denys Séguret, Jan 02 '19 at 15:35
@DenysSéguret yeah no but how would b = a.splice() assign b a copy of a — John Krakov, Jan 02 '19 at 15:36

score 2 · Accepted Answer · answered Jan 02 '19 at 15:16

In the first line you are creating the array and pointing a variable to that array. In the second line let b = a;, here you are pointing b variable to the a array again.

So both a and b variables will be pointing to the same array. When you do a change to the array, both a and b values will be changed.

How come if a = b, then b = a?

1 Answers1