Javascript

How to make a copy of an array in JavaScript

Making copies of arrays in JavaScript is not as straightforward as it seems. It certainly not as easy as b = a. Let’s...

Written by Luci · 2 min read >

Making copies of arrays in JavaScript is not as straightforward as it seems. It certainly not as easy as b = a.

Let’s look at an example:

let a = [1, 2, 3]
const b = a
console.log( a === a)
---
a is array of  [ 1, 2, 3 ]
b is array of  [ 1, 2, 3 ]
Is a equal b?  true

So far so good. Here’s what we did:

  • a is an array.
  • We assigned it to b.
  • Now b is also an array.
  • It contains the same elements as a.
  • And a equals b.

Is the job done? Not quite.

Let’s add one more step: change a value of one of the elements in a array after we “copy” it to b:

let a = [1, 2, 3]
const b = a
a[1] = 3

console.log('a is array of ', a)
console.log('b is array of ', b)
console.log('Is a equal b? ', a === b)
---
a is array of  [ 1, 3, 3 ]
b is array of  [ 1, 3, 3 ]
Is a equal b?  true

Wat? We changed the value of an element of a array, but b also god changed?

That happens because an array is a reference type in JavaScript. When you assign a value of a reference type to another variable, you are not copying the value itself, you just make another reference to it.

In our case:

let a = [1, 2, 3]
const b = a
  • [1, 2, 3] is an array. a is a reference to that array.
  • We make new reference b which is the same as a. That means that both a and b refer to the same array. And that’s why changing the array via a also changes it in b.

The whole topic of value types vs reference types is complicated. If you want to know more about that, leave a comment below (you’ll need to log in for that). And I’ll write something up.

However, sometimes we need to make a real copy of an array, not just make another reference to it. Like if we need to change it in one place without changing in the other.

Making a true copy of an array

There are multiple ways of cloning an array in JavaScript. In my opinion, the simplest one is to take advantage of ES6 spread syntax:

let a = [1, 2, 3]
const b = [...a]
a[1] = 3

console.log('a is array of ', a)
console.log('b is array of ', b)
console.log('Is a equal b? ', a === b)
---
a is array of  [ 1, 3, 3 ]
b is array of  [ 1, 2, 3 ]
Is a equal b?  false

[...a] effectively makes a new array (for real) with all the elements the same as the original array. If we assign it to b, then b will still be a reference, however, it will point to the new array and not the original. Note that a == b is false. That’s how we know that a and b do not refer to the same array any more.

Yes, comparing reference types in JavaScript is not simple either. What to know more? Let me know by commenting under this post.

Cloning a two-dimensional array

Ok, that’s all good. But here’s a trick question. What if we want to clone a two-dimensional array? Is doing [...a] enough?

Not quite. In JavaScript, a two-dimensional array is just an array of arrays. Therefore, cloning one dimension is not enough. We also need to clone all the sub-dimension arrays. Here’s how we do that:

function cloneGrid(grid) {
  // Clone the 1st dimension (column)
  const newGrid = [...grid]
  // Clone each row
  newGrid.forEach((row, rowIndex) => newGrid[rowIndex] = [...row])
  return newGrid
}

// grid is a two-dimensional array
const grid = [[0,1],[1,2]]
newGrid = cloneGrid(grid)

console.log('The original grid', grid)
console.log('Clone of the grid', newGrid)
console.log('They refer to the same object?', grid === newGrid)
---
The original grid [ [ 0, 1 ], [ 1, 2 ] ]
Clone of the grid [ [ 0, 1 ], [ 1, 2 ] ]
They refer to the same object? false

Or if we take advantage of ES6 Array.map operation, we can make cloneGrid function even simpler:

const cloneGrid = (grid) => [...grid].map(row => [...row])

Happy coding!

Written by Luci
I am a multidisciplinary designer and developer with a main focus on Digital Design and Branding, located in Cluj Napoca, Romania. Profile
0 0 votes
Article Rating
Subscribe
Notify of
guest
0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x