Redefine prototype method

Question

I'm trying to simulate the "class" syntax in JavaScript. How do I call the function from an object's prototype when redefining it? In the example, I'm trying to extend the Bear object's sound function.

function Animal(name) {
    this.name = name;
}
Animal.prototype.sound = function() { console.log("I'm " + this.name); }

function Bear(name) {
    Animal.call(this, name)
}
Bear.prototype = new Animal();
Bear.prototype.sound = function() { this.prototype.sound(); console.log("growl!"); }

const cal = new Bear("Callisto")
cal.sound() // Should be: I'm Callisto growl!
console.log(cal)

[Don't use `new Animal` to create the prototype object](https://stackoverflow.com/questions/12592913/what-is-the-reason-to-use-the-new-keyword-here)! — Bergi, Jan 26 '19 at 15:33
So, what then? This: Object.setPrototypeOf(Bear, Animal.prototype); ? — Ondrej Sotolar, Jan 26 '19 at 15:40
Either `Object.setPrototypeOf(Bear.prototype, Animal.prototype)` (which I'd recommend to avoid though) or `Bear.prototype = Object.create(Animal.prototype)`. — Bergi, Jan 26 '19 at 15:46
Could you please give a reason, why create over setPrototypeOf? — Ondrej Sotolar, Jan 26 '19 at 15:49

score 1 · Accepted Answer · answered Jan 26 '19 at 15:35

1

You could directly access the method on the Animals prototype:

Bear.prototype.sound = function() { 
 Animal.prototype.sound.call(this); 
 console.log("growl!"); 
};

answered Jan 26 '19 at 15:35

Jonas Wilms

132,000
20
149
151

Redefine prototype method

1 Answers1