JavaScript inheritance not working as expected

Question

I have my master object:

var Person = function() {
    this.canTalk = true;
}

Method I want all sub-objects to inherit:

Person.prototype.greet = function() {
    if( this.canTalk ) {
        console.log( 'Hi, I am ' + this.name );
    }
}

Sub-object that inherits from Employee

  var Employee = function( name, title) {
        Person.call( this );
        this.name = name;
        this.title = title;
    }

Instantiation:

var robert = new Employee( "Robert Rocha", "Software Developer" );
robert.greet();

greet() gets the error: Uncaught TypeError: robert.greet is not a function

What am I doing wrong?

The Employee prototype is not inheriting from the Person prototype. You are calling the Person constructor, but not getting any of the data from the Person prototype. — jfriend00
– jfriend00, Commented Jun 17, 2015 at 23:27

Daniel A. White · Accepted Answer · 2015-06-17 23:19:50Z

4

You need to extend the prototype of Employee.

Employee.prototype = Object.create(Person.prototype);
Employee.prototype.constructor = Employee;

answered Jun 17, 2015 at 23:19

Daniel A. White

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4 Comments

Robert Over a year ago

I thought that simply assigning a value to the prototype would cause all sub-objects to inherit.

Robert Over a year ago

Ahh ok so I think I understand. The Object.prototype statement is like saying, "I want to extend blah to this current object" so to speak, and the Object.create()` specifies what object it is that you are extending?

Daniel A. White Over a year ago

not really. Object.prototype already exists so I'll use MyFunc.prototype is I want all instances of MyFunc to have these properties. Object.create is merely copying it.

Shashank Over a year ago

@RobertRocha Just fyi, to get the actual prototype of any given object, you do Object.getPrototypeOf(obj).

cade galt · Accepted Answer · 2015-07-11 22:35:24Z

-2

The key thing you need to do is set the prototype chain. You correctly called the parent constructor and passed it the value of this. You can do this very simply:

Employee.prototype = Person.prototype;

However, now when you add a method to Person, Employee will have access to it as well. If you have a special use case this will work but typically you won't want to do this.

Using the more common method, and when you add a method to Employee.prototype it will not be available to Person.

Employee.prototype = Object.create(Person.prototype);

Note that you are still over-writing Employee.prototype and need to define methods after this over-write.

edited Jul 11, 2015 at 22:35

answered Jun 17, 2015 at 23:51

cade galt

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2 Comments

Bergi Over a year ago

It "works", yes, but is a very bad practise - you're not setting a prototype chain here

Bergi Over a year ago

Just for reference: stackoverflow.com/questions/11088365/…

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