JSON object != JavaScript object? [duplicate]

Question

This question already has answers here:

Object comparison in JavaScript [duplicate] (10 answers)

Closed 8 years ago.

For convenience I wrote a simple toJSON prototype, for handling JSON that I know to be safe:

String.prototype.toJSON = function () {
    return JSON.parse(this.valueOf());
};

I am using it in testing my web-services. Unfortunately even with this simple test:

var v0 = '{"echo":"hello_world"}'.toJSON(), v1 = {"echo": "hello_world"};

It fails:

console.log(v0 == v1);            // false
console.log(v0 === v1);           // false
console.log(v0.echo == v1.echo);  // true
console.log(v0.echo === v1.echo); // true

What do I not know about JavaScript which is causing this issue?

Answer 1

An object in JavaScript, just like everything else except primitives(int, string, Boolean) is a reference.

Having 2 different duplicate objects, means having 2 different references that point to different places within the hype.

You can implement something as simple as that, to basically iterate over all of the primitive properties of an object, and compare them one by one:

Object.prototype.equals = function(x)
{
    for(p in this)
    {
        switch(typeof(this[p]))
        {
            case 'object':
                if (!this[p].equals(x[p])) { return false }; break;
            case 'function':
                if (typeof(x[p])=='undefined' || (p != 'equals' && this[p].toString() != x[p].toString())) { return false; }; break;
            default:
                if (this[p] != x[p]) { return false; }
        }
    }

    for(p in x)
    {
        if(typeof(this[p])=='undefined') {return false;}
    }

    return true;
}

Answer 2

Just because you have the same content, that does not mean that you have the same object instance. If you had done v1 = v0 instead of initializing v1 seperatly the first two would have returned true.

Update: If you need to compare the two instances to find if they have equal content then you need to define a function which compares each member.

Answer 3

For two objects to be == they must be the same instance. v0 and v1 are two different instances. If you want to do a deep comparison of objects you can use something like underscore's isEqual method: http://underscorejs.org/#isEqual

_.isEqual(v0, v1);