How to make comparison of objects `a == b` to be true?

Question

Here is one of the questions in JavaScript online-test before job interview:

function F(){};

var a = new F();
var b = new F();

Q: How to make comparison a == b to be true? (e.g. console.log(a == b) // true)

I answered that it's impossible because a and b are two different instances of F and equal comparison in JS in case of non-primitives compares reference.

But some time ago I've read article "Fake operator overloading in JavaScript" by Axel Rauschmayer: http://www.2ality.com/2011/12/fake-operator-overloading.html — and I wonder if there is a hack to fake operator overload in comparison of objects?

Answer 1

It really depends on what they mean by "How to make comparison a == b to be true?"

If you're allowed to change the constructor, then you could make your constructor a singleton:

function F(){
    if (!F.instance) {
        F.instance = this;
    } else {
        return F.instance;
    }
};
var a = new F();
var b = new F();
if (a === b) {
    //they are the same
}

If they want you to keep everything as it is but have a comparision that contains a == b then you could write the following:

if ("" + a == b) {
}

If they want to know methods of determine whether the two objects are instances of the same constructor function, then you could compare the constructor property or the __proto__ property:

if (a.constructor === b.constructor) {
}

if (a.__proto__ === b.__proto__) {
}

If they want to know methods of dermine whether these two objects have the same properties, you can either compare their JSON string:

if (JSON.stringify(a) === JSON.stringify(b)) {
}

or you write a function that recursively compares all the properties in both objects (deep comparision).

And the most simple answer to the question "How to make comparison a == b to be true?":

var a = new F();
var b = new F();

b = a;

if (a === b) {
    //surprise!!!
}