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I'm trying to use prototypal inheritance, but I'm having trouble.

This doesn't work

var Parent = function(){
}

var Child = function(){
    this.__proto__ = new Parent();
}

var child = new Child();

console.log(child instanceof Child) //logs false

But this does

var Parent = function(){
}

var Child = function(){

}
Child.prototype = new Parent();

var child = new Child();

console.log(child instanceof Child) // logs true

The only reason why I want the first option is so that I can leverage the parent's constructor. I'm guessing this is the problem, but I'm not that great at javascript. How do I make this work?

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5
  • When you say this.__proto__ = new Parent() you're saying "Okay new object, stop being a Child and start being a Parent instead," so it's not surprising that the object is no longer and instance of Child. Commented Nov 27, 2013 at 15:47
  • @apsillers is that because of this? Why doesn't the second method do the same thing then? Is it because I am not in the scope of new Child when I set the prototype? Commented Nov 27, 2013 at 15:50
  • Ah, I understand your confusion now; I'll update my answer. Commented Nov 27, 2013 at 15:54
  • 1
    Hi Josh, it's a little late but maybe the following answer can give you a better understanding of prototype (shared members) and this (instance members): stackoverflow.com/a/16063711/1641941 Commented Nov 27, 2013 at 23:51
  • but this is actually Classical Inheritance not Prototypal Inheritance. Commented Mar 31, 2017 at 12:31

1 Answer 1

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2

The better way to do this is to call the Parent constructor on this:

var Child = function(){
    Parent.call(this);
}

That way, the Parent constructor code runs with its this set to the this in the Child constructor, but you don't change the __prototype__ of this.

Your two examples do produce a child instance that is structurally the same. However, the main difference is that in your first example, Child.prototype != child.__proto__. Although it is true that Child.prototype and child.__proto__ are both objects with a __proto__ of Parent.prototype, they are not the exact same object, so instanceof fails.

You may also want to do Child.prototype = Object.create(Parent.prototype); so that Child instances have access to Parent's prototype method. (Currently you don't have any methods on Parent.prototype, but maybe you will someday.)

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7 Comments

It is entirely my intention to have common methods on the Parent or its prototype
Don't I need to set the result of Parent.call(this) to something on Child? And, does Child.prototype = Object.create(Parent.prototype); replace Child.prototype = new Parent();?
@JoshC. No, Parent.call(this) stands on its own, because it mutates this. The Parent constructor does things like this.foo = 5, so after Parent.call(this) completes, your this has a foo property.
@JoshC. Yes, you should do Child.prototype = Object.create(Parent.prototype); instead of Child.prototype = new Parent(); Here's an example of why: stackoverflow.com/a/20201556/710446. Basically, the Parent constructor might do things per-instance (like set a unique ID), and you don't want all your Child instances sharing a single instance ID.
@JoshC. Yes, that's the correct way to use call. You could also use apply to pass in an array of arguments, like Parent.apply(this, [arg1, arg2, arg3]). This is especially handy if you want to pass in all of Child's arguments to Parent, by doing Parent.call(this, arguments).
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