Question

Note: this isn’t one of the millions dups of the common array copy ‘problem’ where using arr.slice(0) fixes this ‘problem’

That said, I want to understand why I am getting this unexpected result:

var oldArr = [[1,2],[3,4]];
var find = oldArr[1];

var newArr = oldArr.slice(0);
console.log(newArr.indexOf(find)); //1?

//proof that newArr is NOT referenced to oldArr
newArr[0] = "Hi";
newArr[1] = "How are you?";
console.log(oldArr+" "+newArr); //"1,2,3,4 Hi,How are you?"

jsFildde Demo

If you replace find with any of the following alternatives, it returns the expected -1:

• Use [3,4] directly
• Use a variable holding [3,4]
• Use a variable with reference of another array holding [3,4]

I can’t find any explanation on why there is any difference between these last three methods and the first example. As far as I know, there shouldn’t be any.

Any ideas?

Answer 1

In:

[[1,2],[3,4]]

There are three array objects created.

Only the outer one is being slice‘d. This results in a “shallow copy”.

Consider this:

var a = [1,2]
var b = [3,4]
var c = [a,b]
var d = c.slice(0)
d[0] === a       // true, which means it is the /same/ object
d[0][0] = "Hi!"
a                // ["Hi!", 2]

Happy coding!