Does javascript Array.slice(0) return a shallow copy?

Question

IIUC Array.slice(0) returns a copy of the array. Is it a shallow copy? In other words the array items still have the same memory location, but the array container gets assigned a new one?

Effectively:

let oldArray = ['old', 'array'];
let newArray = oldarray.slice(0);
let same = oldArray[0] === newArray[0]; //true 
let same = oldArray === newArray; //false

Yes, if the items in the array are objects, they will not be coped, only referenced in a new array container. You have to go to *some significant effort* to actually deep copy arrays and objects in general (thankfully). — CertainPerformance, Apr 18 '18 at 02:10

score 5 · Accepted Answer · answered Apr 18 '18 at 02:11

5

Yes,see demo:

var o = [{
  x: 1
}]
var o2 = o.slice(0)
o2[0].x = 2
console.log(o[0].x)

answered Apr 18 '18 at 02:11

xianshenglu

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score 3 · Answer 2 · answered Apr 18 '18 at 02:16

You are correct - it does create a shallow copy.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice

The slice() method returns a shallow copy of a portion of an array into a new array object selected from begin to end (end not included). The original array will not be modified.

If you are wanting a deep copy you can do the ye ole dirty dirt:

var arr = [1]; 
var arr2 = JSON.parse(JSON.stringify(arr));

arr[0] = 99;
arr2[0] = 1000;

console.log({arr, arr2});

Does javascript Array.slice(0) return a shallow copy?

2 Answers2