Question

Why is it that just by using call you’re adding properties to the Employee instance named jc?

I’m wondering why jc.hasOwnProperty('firstName'); results to true.

I didn’t even do prototype inheritance yet. This is all the code, no more no less:

I understand that call just changes the this while passing in arguments, but to add properties? I don’t know what’s going on…

function Person(firstName, lastName){
  this.firstName = firstName;
  this.lastName = lastName;
}
function Employee(firstName, lastName, position){
  Person.call(this,firstName,lastName);
  this.position = position;
}
var jc = new Employee('JC','Viray','Developer');
jc.hasOwnProperty('firstName'); //true

UPDATE

I get it now.. the solution to my problem is:
stop thinking in C#/Java mentality.. I lost track that Person is still a FUNCTION despite it being a “constructor type function”… -_- once a function, still a function..

Answer 1

I’m wondering why jc.hasOwnProperty(‘firstName’); results to true.

Because of this line in the Person function:

this.firstName = firstName;

When you say

Person.call(this, firstName, lastName);

You’re actually calling the Person function, and telling Person to set its this value to what’s provided in the first argument—in your case, the Employee object you’re in the process of building. The subsequent arguments to call then get passed along as regular parameters.

As a result, Person gets called with this set to be your current Employee object, and picks up a firstName and lastName property.

---

Or consider an alternate example.

var obj = { };
Person.call(obj, "Adam", "Rackis");
obj.hasOwnProperty('firstName'); //still true