Javascript: operator overloading

Solution 1:

As you've found, JavaScript doesn't support operator overloading. The closest you can come is to implement toString (which will get called when the instance needs to be coerced to being a string) and valueOf (which will get called to coerce it to a number, for instance when using + for addition, or in many cases when using it for concatenation because + tries to do addition before concatenation), which is pretty limited. Neither lets you create a Vector2 object as a result. Similarly, Proxy (added in ES2015) lets you intercept various object operations (including property access), but again won't let you control the result of += on Vector instances.

For people coming to this question who want a string or number as a result (instead of a Vector2), though, here are examples of valueOf and toString. These examples do not demonstrate operator overloading, just taking advantage of JavaScript's built-in handling converting to primitives:

valueOf

This example doubles the value of an object's val property in response to being coerced to a primitive, for instance via +:

function Thing(val) {
    this.val = val;
}
Thing.prototype.valueOf = function() {
    // Here I'm just doubling it; you'd actually do your longAdd thing
    return this.val * 2;
};

var a = new Thing(1);
var b = new Thing(2);
console.log(a + b); // 6 (1 * 2 + 2 * 2)

Or with ES2015's class:

class Thing {
    constructor(val) {
      this.val = val;
    }
    valueOf() {
      return this.val * 2;
    }
}

const a = new Thing(1);
const b = new Thing(2);
console.log(a + b); // 6 (1 * 2 + 2 * 2)

Or just with objects, no constructors:

var thingPrototype = {
    valueOf: function() {
      return this.val * 2;
    }
};

var a = Object.create(thingPrototype);
a.val = 1;
var b = Object.create(thingPrototype);
b.val = 2;
console.log(a + b); // 6 (1 * 2 + 2 * 2)

toString

This example converts the value of an object's val property to upper case in response to being coerced to a primitive, for instance via +:

function Thing(val) {
    this.val = val;
}
Thing.prototype.toString = function() {
    return this.val.toUpperCase();
};

var a = new Thing("a");
var b = new Thing("b");
console.log(a + b); // AB

Or with ES2015's class:

class Thing {
    constructor(val) {
      this.val = val;
    }
    toString() {
      return this.val.toUpperCase();
    }
}

const a = new Thing("a");
const b = new Thing("b");
console.log(a + b); // AB

Or just with objects, no constructors:

var thingPrototype = {
    toString: function() {
      return this.val.toUpperCase();
    }
};

var a = Object.create(thingPrototype);
a.val = "a";
var b = Object.create(thingPrototype);
b.val = "b";
console.log(a + b); // AB

Solution 2:

As T.J. said, you cannot overload operators in JavaScript. However you can take advantage of the valueOf function to write a hack which looks better than using functions like add every time, but imposes the constraints on the vector that the x and y are between 0 and MAX_VALUE. Here is the code:

var MAX_VALUE = 1000000;

var Vector = function(a, b) {
    var self = this;
    //initialize the vector based on parameters
    if (typeof(b) == "undefined") {
        //if the b value is not passed in, assume a is the hash of a vector
        self.y = a % MAX_VALUE;
        self.x = (a - self.y) / MAX_VALUE;
    } else {
        //if b value is passed in, assume the x and the y coordinates are the constructors
        self.x = a;
        self.y = b;
    }

    //return a hash of the vector
    this.valueOf = function() {
        return self.x * MAX_VALUE + self.y;
    };
};

var V = function(a, b) {
    return new Vector(a, b);
};

Then you can write equations like this:

var a = V(1, 2);            //a -> [1, 2]
var b = V(2, 4);            //b -> [2, 4]
var c = V((2 * a + b) / 2); //c -> [2, 4]

Solution 3:

It's possible to do vector math with two numbers packed into one. Let me first show an example before I explain how it works:

let a = vec_pack([2,4]);
let b = vec_pack([1,2]);

let c = a+b; // Vector addition
let d = c-b; // Vector subtraction
let e = d*2; // Scalar multiplication
let f = e/2; // Scalar division

console.log(vec_unpack(c)); // [3, 6]
console.log(vec_unpack(d)); // [2, 4]
console.log(vec_unpack(e)); // [4, 8]
console.log(vec_unpack(f)); // [2, 4]

if(a === f) console.log("Equality works");
if(a > b) console.log("Y value takes priority");

I am using the fact that if you bit shift two numbers X times and then add or subtract them before shifting them back, you will get the same result as if you hadn't shifted them to begin with. Similarly scalar multiplication and division works symmetrically for shifted values.

A JavaScript number has 52 bits of integer precision (64 bit floats), so I will pack one number into he higher available 26 bits, and one into the lower. The code is made a bit more messy because I wanted to support signed numbers.

function vec_pack(vec){
    return vec[1] * 67108864 + (vec[0] < 0 ? 33554432 | vec[0] : vec[0]);
}

function vec_unpack(number){
    switch(((number & 33554432) !== 0) * 1 + (number < 0) * 2){
        case(0):
            return [(number % 33554432),Math.trunc(number / 67108864)];
        break;
        case(1):
            return [(number % 33554432)-33554432,Math.trunc(number / 67108864)+1];
        break;
        case(2):
            return [(((number+33554432) % 33554432) + 33554432) % 33554432,Math.round(number / 67108864)];
        break;
        case(3):
            return [(number % 33554432),Math.trunc(number / 67108864)];
        break;
    }
}

The only downside I can see with this is that the x and y has to be in the range +-33 million, since they have to fit within 26 bits each.