Javascript Classical Inheritance calling parent

Question

I want to call the super method in an extended javascript 'class' by applying classical inheritance.

function Person(name, age) {
    this._name = name;
    this._age = age;
}
Person.prototype.exposeInfo = function() {
    alert(this._name + ' - ' + this._age);    
}

function Employee(name, age) {
    this.parent.constructor.call(this, name, age);
}
Employee.prototype.exposeInfo = function() {
    alert('Call employee');
    this.parent.exposeInfo();    
}

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


var p1 = new Person('John Doe', 30);
p1.exposeInfo();

var p2 = new Employee('John Foobar (empl.)', 35);
p2.exposeInfo();

JS Fiddle

The problem is that the method is not being called in the extended class, but only in the parent (Person).

Answer 1

That's because the overriding exposeInfo is being attached to the former prototype object, which is then replaced:

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

You'll want to reverse the order, attaching methods after creating the prototype:

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

Employee.prototype.exposeInfo = function() {
    // ...
}

---

You'll also need to use .call() or .apply() with exposeInfo as you did with the constructor:

Employee.prototype.exposeInfo = function() {
    alert('Call employee');
    this.parent.exposeInfo.apply(this, arguments);    
}

Otherwise, the value of this will be determined by the last member operator:

// so, calling:
this.parent.exposeInfo();

// is equivalent to:
alert(this.parent._name + ' - ' + this.parent._age);