Javascript unbind variables

Question

I have the following code that copy variable value then change its value.

var a = [{name: 'John Doe'}];
var b = a;

document.write(b[0].name + '<br />');
document.write(a[0].name + '<br /><br />');

b[0].name = 'Jane Doe';

document.write(b[0].name + '<br />');
document.write(a[0].name + '<br />');

But somehow that also change first variable value

How do I make variable A to keep its value?

You have to deeply clone `a`, one way to do it is `var b = JSON.parse(JSON.stringify(a))` — Mauricio Poppe, Jun 01 '16 at 22:08
Possible dupe of: http://stackoverflow.com/questions/15722433/javascript-copy-array-to-new-array — Oisin, Jun 01 '16 at 22:08
Variable a is being passed into b by reference. Since a and b have the same reference, whenever you change anything on one it also changes the other. To prevent this you'll need to do a deep copy of a onto b, copying all of the individual properties. — nurdyguy, Jun 01 '16 at 22:08
Objects (including Arrays) are assigned by reference in JavaScript. You must manually copy them if you need to maintain two separate `.name` values. — Mulan, Jun 01 '16 at 22:09
Sorry if its duplicated, I dont know the correct terms when I search the answer before :( — ilmansg, Jun 01 '16 at 22:24

score 1 · Accepted Answer · edited Jun 20 '20 at 09:12

You are just assigning the a-reference to b. What you may want to do is copying the array so that changes to a are not reflected to b. See this thread for the different assignments in JavaScript.

Take also a look at fastest way to duplicate an array. It's about the speed of different array copy methods. Quick answer is:

var b = a.slice();

The slice() method creates a new array with the same elements than the original one. But be careful it's only a shallow copy.

You could also use the JSON.stringify() and JSON.parse()methods for doing a deep copy.

Javascript unbind variables

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