Strange behavior array in object JavaScript

Question

Why b.first[0] return "t" and how I can avoid this?

I need safe "q" in b.first[0]

var extend = function(o,p){
    for(prop in p){
        o[prop] = p[prop];
    }
    return o;
};

var a = {first:['q','w']};
var b = {};

extend(b,a);

document.write(a.first[0]); //q
document.write(b.first[0]); //q

a.first[0] = 't';

document.write(a.first[0]); // t
document.write(b.first[0]); // t ?????????????????????

Because a.first is a reference to an array. When assigning b.first = a.first (which is what extend seems to do) then b gets a reference to the same array. — RobG
– RobG, Commented Apr 5, 2013 at 5:57
You need a deep copy of your object. stackoverflow.com/questions/122102/… — yunzen
– yunzen, Commented Apr 5, 2013 at 6:02
In the real I use some of deepExtend, but it seems my version is not good. I will try other versions. Thanks! — Aleksey
– Aleksey, Commented Apr 5, 2013 at 6:35

Community · Accepted Answer · 2017-05-23 10:25:12Z

5

This is an issue relating to the concept that extending b by a doesn't recreate the data from a. If some of the data is an object (like an Array), it instead just "points" to that array instead of creating a new, identical array. Essentially you're storing two pointers to the same array, so when you change one, you change the other.

Here is a answer which discusses the idea of "cloning" an object in Javascript in more detail.

https://stackoverflow.com/a/728694/1570248

edited May 23, 2017 at 10:25

CommunityBot

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answered Apr 5, 2013 at 5:53

Abe Haskins

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