Problem with .sort() method in Javascript, two arrays sort instead of only one

Question

I'm trying to define an array b to be an independent copy of an array argument a. Thus, I want to change b without changing a.

But here is a mystery: When I sort b, a is sorted as well!

function firstDuplicate(a) {
    let b = a;

    console.log(`this is a : ${a}`)  // this is a : 2,1,3,5,3,2 
    console.log(`this is b : ${b}`)  // this is b : 2,1,3,5,3,2

    b.sort()                    
    console.log(`this is a : ${a}`)  // this is a : 1,2,2,3,3,5 (my problem)
    console.log(`this is b : ${b}`)  // this is b : 1,2,2,3,3,5
}

firstDuplicate([2, 1, 3, 5, 3, 2])

How can I avoid this?

Thank you so much for your replies.

they share the reference, try copying the array and sort, it's not a problem with the sort(). — Tinu Jos K
– Tinu Jos K, Commented Feb 11, 2020 at 13:43
It happens because of 'shallow copy'. let b = JSON.parse(JSON.stringify(a)); Would do very deep copy. Actually, created a new instance of the same information... — ghchoi
– ghchoi, Commented Feb 14, 2020 at 3:12

Michael Bianconi · Accepted Answer · 2020-02-11 13:42:47Z

4

You aren't duplicating a, you're just creating a new reference to it. Copy the array with slice(), then sort it.

let b = a.slice().sort();

answered Feb 11, 2020 at 13:42

Michael Bianconi

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bryan60 · Accepted Answer · 2020-02-11 13:45:30Z

0

Arrays are stored by reference in javascript, so when you do

 let b = a

You’re not duplicating, you’re just setting b and a to point at the same array. This would work:

 let b = [...a]

This declares a new array and uses the spread operator to fill it with the contents of a

But be careful because if objects or arrays are inside your original array, the two different arrays are still pointing at the same objects or arrays.

answered Feb 11, 2020 at 13:45

bryan60

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Tinu Jos K · Accepted Answer · 2020-02-11 13:47:00Z

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by doing

let a = b;

you are copying the reference, so any change made to the either of the array will reflect in the other array.

here you have to do a copy,

let a = [...b];

But using the spread operator to copy the array will create a shallow copy, if you want to do a deep copy of the array.

let a = JSON.parse(JSON.stringify(b));

and then perform sort on it

answered Feb 11, 2020 at 13:47

Tinu Jos K

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Sam Anderson · Accepted Answer · 2020-02-11 13:51:29Z

As mentioned, in Javascript the line let b = a is essentially saying make b point to the same array as a. This is because Arrays are stored by reference. As a result any modifications made to a will also be applied to b.

The solution is to create a new array and let b point to this. This can be done multiple ways

let b = Array.from(a)
let b = [...a]
let b = a.slice().join('')

Each of these returns a new Array rather than a reference to the original.

To prove this you could run the following

let a = [1,2,3];
let b = a;
let c = Array.from(a);

a === b  // true
a === c  // false (because it is a new value)

a[2] = 'new';   // [1,'new',3]

a === b // true (because b also points to [1,'new',3])
a === c // false (c still equals [1,2,3] and is still pointing to a different value)

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