Typescript polymorphism

Question

Please have a look on this code:

class Greeter {
    greeting: string;
    constructor(message: string) {
        this.greeting = message;
    }
    greet() {
        return "Hello, " + this.greeting;
    }
}

class Ge extends Greeter {
    constructor(message: string) {
        super(message);
    }
    greet() {
        return "walla " + super.greet();
    }
}

let greeter = new Ge("world");

console.log(greeter.greet()); // walla Hello, world
console.log((<Greeter> greeter).greet()); // walla Hello, world

I would expect the second log to print Hello, world. Looking at the transpiled Javascript code, I see the exact same command so it's not that a surprise.

The real question is, how do you cast greeter to its extended class?

Answer 1

Methods in JS (and TS) are attached to a prototype which is attached to each instance (something like a virtual method table). When you call a method, the actual function is retrieved from the instance's prototype chain rather than the known type of the object. The closest equivalent I'm aware of are virtual methods in C++.

In code:

let greeter = new Ge('world');
// You expect:
Greeter.prototype.greet.call(greeter);
// JS actually does:
greeter.prototype.greet.call(greeter);

Answer 2

You already did cast greeter to it's parent class.

An overriden method in a class doesn't change behavior when cast as its parent.