How to use Array.fill() with the same object but different copies for every index?

Question

I'm creating an array and populating it using the fill method but changing array1[0].foo changes all objects in the array.

const array1 = Array(2).fill({ foo: null })
array1[0].foo = 'bar' // [ { foo: 'bar' }, { foo: 'bar' } ]

Is there a way to use fill with different copies of the same object for every index?

You mean different copies. And, no. – SLaks May 31 '18 at 17:20 — SLaks, May 31 '18 at 17:20

Nina Scholz · Accepted Answer · 2018-05-31T17:24:03.597

1

It does not work with Array#fill, because it uses a constant value.

You could use Array.from and map the object.

const array = Array.from({ length: 2 }, _ => ({ foo: null }));

array[0].foo = 'bar';

console.log(array);

edited May 31 '18 at 17:24

answered May 31 '18 at 17:22

Nina Scholz

Wow, that did work, thanks. To complement, I was cloning an array in the first place so it ended up `Array.from(oldArray, () => ({ foo: null }))`. – Ramon Balthazar May 31 '18 at 17:28
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@Razär: Then you might as well call `.map()`. – SLaks May 31 '18 at 17:31

1 Answers1