How javascript prototypical inheritance works for the Array, function

Question

I need to understand how the Array object gets a reference to Object.

For example, when I create var arr = []; it has Array.prototype ---> Object.prototype ---> null.

I want to achieve above for the example below:

Suppose I have a function xyz(), where xyz.prototype.somefunction = function() { }.

I have another function abc(), where abc.prototype.anotherfunction = function() { }.

When I create an object of abc() — as in var obj = new abc() — I want it to have a prototype chain like obj ---> abc.prototype ---> xyz.prototype ---> object.prototype ---> null.

Please suggest the best way to do this.

What are you trying to achieve? Or is this some kind of academic learning exercise? — user663031
– user663031, Commented Jun 16, 2017 at 16:49
@torazaburo is it possible when i create an object of abc same time its does what mentioned above. Right now i need to add the Object.setPrototyeof() se — vikas bisht
– vikas bisht, Commented Jun 16, 2017 at 17:13

Ben Aston · Accepted Answer · 2017-06-16 19:07:58Z

1

function xyz() {}
function abc() {}
var p = new xyz();
abc.prototype = p;

var o = new abc();
o.__proto__ === p // true
o.__proto__.__proto__ === xyz.prototype // true
o.__proto__.__proto__.__proto__ === Object.prototype // true
o.__proto__.__proto__.__proto__.__proto__ === null // true

Or:

function xyz() {}
function abc() {}
var p = Object.create(xyz.prototype);
abc.prototype = p;    

var o = new abc();
o.__proto__ === p // true
o.__proto__.__proto__ === xyz.prototype // true
o.__proto__.__proto__.__proto__ === Object.prototype // true
o.__proto__.__proto__.__proto__.__proto__ === null // true

Or:

class xyz {}
class abc extends xyz {}

var o = new abc();
o.__proto__.__proto__ === xyz.prototype // true
o.__proto__.__proto__.__proto__ === Object.prototype // true
o.__proto__.__proto__.__proto__.__proto__ === null // true

Or:

function xyz () {}
const abc = {
    __proto__: Object.create(xyz.prototype)
}

abc.__proto__.__proto__ === xyz.prototype // true
abc.__proto__.__proto__.__proto__ === Object.prototype // true
abc.__proto__.__proto__.__proto__.__proto__ === null // true

Or:

function xyz () {}
function abc() {}
Object.setPrototypeOf(abc.prototype, xyz.prototype);

var o = new abc();
o.__proto__.__proto__ === xyz.prototype // true
o.__proto__.__proto__.__proto__ === Object.prototype // true
o.__proto__.__proto__.__proto__.__proto__ === null // true

edited Jun 16, 2017 at 19:07

answered Jun 16, 2017 at 17:14

Ben Aston

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2 Comments

castletheperson Over a year ago

The one with const abc = { isn't declaring a function like the rest.

vikas bisht Over a year ago

Great. This what i was looking for .. :)

gauravmuk · Accepted Answer · 2017-06-16 16:50:05Z

0

You can do prototypical inheritance:

abc.prototype = new xyz();

now, when you create a new object, it will follow the chain you desire.

answered Jun 16, 2017 at 16:50

gauravmuk

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