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

Question

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Object comparison in JavaScript [duplicate] (10 answers)

Closed 8 years ago.

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!!!
}

Answer 2

my best answer would be this since you can compare different functions:

console.log(a.constructor+"" === b.constructor+"");

as it returns the functions as strings and then compare them literally .

example test:

function f1(){}
function f2(){}
var a = new f1(),
    b= new f2();
console.log(a.constructor+"" === b.constructor+"");
b = new f1();
console.log(a.constructor+"" === b.constructor+"");

DEMO

note: the === sign is not needed as the third would be for type comparison and both are strings at that point so using == would do exactly the same thing

EDIT: my actual answer to the question however would be: by removing new from the initialization