Question

Func2 is called when I click on button . Why i don’t get any popup? Should I see both alert alert (“override1”) and alert (“override2”) after the first one?

// JavaScript Document
function person(name, surname) {
    this.name = "";
    this.surname = "";
    this.age = "11";
    this.setName(name);
    this.setSurname(surname);
    //alert('Person instantiated'); 
}
person.prototype.setName = function(name) {
    this.name = "Sir1" + name;
}
person.prototype.setSurname = function(surname) {
    this.surname = "-" + surname;
}
person.prototype.setAge = function(newAge) {
    this.age = newAge;
}
person.prototype.show = function() {
    alert("override1");
}
function employee(name, surname, company) {
    //if (arguments[0] === inheriting) return;
    person.call(this, name, surname); // -> override del costruttore
    //this.name = "Sir2"+name;
    this.company = company;
};
employee.prototype.show = function() {
    person.prototype.show;
    alert("override2");
}
function test2() {
    employee.prototype = new person();
    // correct the constructor pointer because it points to Person  
    employee.prototype.constructor = employee;
    // Crea un oggetto impiegato da persona
    impiegato = new employee("Antonio", "Di Maio", "Consuldimo");
    //impiegato.show();
    impiegato.show();
}​

Thanks

Answer 1

In test2() you’re replacing the entire employee.prototype with an instance of person, thus overwriting the employee.prototype.show function you’ve defined previously with the one inherited from person. Also, as stated in the answer by codebox, in employee.prototype.show() you’re not calling person.prototype.show(), but merely evaluating it in void context, which has no effect at all.

You’ll have to set employee‘s parent before you define any additional methods on its prototype:

employee.prototype = new person();
employee.prototype.constructor = employee;
employee.prototype.show = function() { ... }

Also, when you call your parent’s method, you need to supply the correct context yourself:

person.prototype.show.call(this);