javascript inheritance

Question

I know there is a lot of similar questions are tons of great answers to this. I tried to look at the classical inheritance methods, or those closure methods etc. Somehow I c

Answer 1

It is not a hack as such; JavaScript is a prototyped language, as defined by Wikipedia as where:

..classes are not present, and behavior reuse (known as inheritance in class-based languages) is performed via a process of cloning existing objects that serve as prototypes.

As it says, classes are not used in JavaScript; each object that you create is descended from the JavaScript Object; all objects in JavaScript have the prototype object, and all instances of objects you create 'inherit' methods and properties from their object's prototype object. Take a look at the MDC prototype object reference for more information.

As of this, when you call the line:

CHILDClass.prototype = new PARENTClass();

This allows the CHILDClass object to add methods and properties to its prototype object from the PARENTClass object, which creates an effect similar to the idea of inheritance present in class-based languages. Since the prototype object affects every instance created of that object, this allows the parent object's methods and properties to be present in every instance of your child object.

If you want to call your parent class's constructor in your child class's constructor, you use the JavaScript call function; this allows you to call the parent class's constructor in the context of the child class's constructor, therefore setting the newly prototyped properties in your child class to what they are set as in the parent class.

You also do not need to put anything where you have specified the *1, since that line is merely used to add the methods and properties to the child class's prototype object; however, bear in mind that it calls the parent class's constructor, so if there are any arguments that are fundamental in the operation of the parent class constructor, you should check that these are present so as to avoid JavaScript errors.