CODEWITHSUNDEEP
javascriptarraysecmascript-6
Peter Toth
Peter Toth

Reputation: 770

Pushing array into array then handling elements inside

so I just came across this, but could anyone explain it to me, why is this phenomenon present here? I didn't find any help online...

1.So if I were to get the following two arrays:

let a = [];
let b = ['','',''];

2.and then push array b into array a three times:

a.push(b);
a.push(b);
a.push(b);

3.but then when I try to change the first element of the first array like so:

a[0][0] = 'foo';

4.then all the first element of ALL the arrays change, not just the first one, as console.log(a) outputs:

(3) [Array(3), Array(3), Array(3)]
0: Array(3)
  0: "foo"
  1: ""
  2: ""

1: Array(3)
  0: "foo"
  1: ""
  2: ""

2: Array(3)
  0: "foo"
  1: ""
  2: ""

I would like to ask, why does this happen and in what way could I change an element based on its index inside of array a

Upvotes: 0

Views: 42

Answers (3)

JBallin
JBallin

Reputation: 9807

b is a pointer to a mutable array, changing the array it points to will cause the other places you point to it to update as well.

You would either push copies of b into the array or create a modified copy of b when trying to mutate the value it points to.

Upvotes: 2

MarketMan39
MarketMan39

Reputation: 208

if you console.log(b), you'll see that as well.
Try doing b[1] = "test" then check the value of a and you'll see all the b's that were pushed into there have test in them too.
b is a reference so anytime you change the value of it, you are altering the places it is referenced to.

Upvotes: 1

tom
tom

Reputation: 1493

Because you're passing b which is just a reference to an object. To have the behavior you want, change your code to be like this:

a.push(b.map((x) => x))

For more see this page.

Upvotes: 1

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