calling a method using prototype

Question

<script> 
 var Kevin = function(){ 
 this.name = 'kevin'
 } 
 Kevin.prototype.getKevin = function(){ 
 alert(this.name); 
 } 
 Kevin.prototype.getKevin();
 function John(){
 this.name = 'john'
 } 
 John.getStaticJohn = function(){
 alert(this.name); 
 }
 John.prototype.getJohn();
 John.getStaticJohn();
</script>

1. Why is that i am getting undefined in both the cases when calling the method using prototype.
2. When i try to call the static method in John class, it prints the output perfectly.

• Brad Christie: Object.prototype just allows you to extend the object from outside of it. Calling it still relies on an instance of said object being defined first.

Answer 1

If you wanted to call the methods from the constructor, you would need to create an anonymous instance:

(new Kevin).getKevin(); // or new Kevin().getKevin()

Answer 2

You're getting undefined because the prototype has no "name" property. Note also that your call to "getStaticJohn()" does not in fact "work perfectly" - it alerts "John" with a capital "J" because it's accessing the "name" property of the function object "John".

When you call a method via an expression of the form something.functionName, then the value of this inside the function will always be the value of something. Thus when you call

John.prototype.getJohn();

the value of this inside the "getJohn()" function will be John.prototype, and not any instance constructed by the "John()" constructor.

If you add this:

John.prototype.name = "John's prototype";

then your call to John.prototype.getJohn() will alert something other than undefined.

• Kevin: Can you lead me to a nice article on prototype where i can understand the concept fully.