Don't change the main object when adding a variable to it in JavaScript

Question

I need to use a variable globally declared, but modify it for the use of a function:

var example = {
    selected: '0',
    list: {
        1: {
            value: '1',
            name: "example 1"
        },
        2: {
            value: '2',
            name: "example 2"
        },
        3: {
            value: '3',
            name: "example 3"
        }
    }
};

window.load(function () {
    var example2 = example;
    example2.info = "newinfo";
    // Use example 2
})

The problem is, after that, if I console.log(example), it contains the "info" variable, absolutely unwanted. I don't even see why it would have it, I purposely defined a new variable to avoid this.

I guess you're looking for prototype https://developer.mozilla.org/en-US/docs/Web/JavaScript/Inheritance_and_the_prototype_chain — enguerranws, Nov 10 '15 at 14:24
See: http://stackoverflow.com/questions/728360/most-elegant-way-to-clone-a-javascript-object — Biketire, Nov 10 '15 at 14:24
Read [Most elegant way to clone javascript object](http://stackoverflow.com/questions/728360/most-elegant-way-to-clone-a-javascript-object) — pistou, Nov 10 '15 at 14:24
view this question for explainantion as to why: http://stackoverflow.com/questions/518000/is-javascript-a-pass-by-reference-or-pass-by-value-language — uesports135, Nov 10 '15 at 14:26
You are assigning objects by reference, so when you change example or example2 they will change the same reference. You have to clone the object then change it. — Ahmad Harb, Nov 10 '15 at 14:30

Nina Scholz · Accepted Answer · 2015-11-10T14:26:55.227

4

The single assignment keeps the reference to the original object. So any changes are reflected to the original object.

Just use JSON.stringify and JSON.parse for a copy.

var example2 = JSON.parse(JSON.stringify(example));

edited Nov 10 '15 at 14:26

answered Nov 10 '15 at 14:24

Nina Scholz

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Don't change the main object when adding a variable to it in JavaScript

1 Answers1