Thread: Simplest Problem but i am unable to understand it, WHY???

Well, it is the simplest OOP concept, i also know that it happens like this, but i am still confused, since i read it the very first time. I like to post this here. May be someone could provide me such a logic that could satisfy me.

PROBLEM:

We know that a super class's object can refer to a sub class object. i.e.

// Class A
class A{
        private int a;
        private int b;
        A(){
                a=0;
                b=0;
            }
            A(int a, int b){
                    this.a=a;
                    this.b=b;
            }
            public show(){
                System.out.println(a+","+b);
            }
 
    }

// Class B
class B extends A{
        private int c;
        B(){
                super();
                c=0;
            }
            B(int a, int b, int c){
                    super(a,b);
                    this.c=c;
            }
            public show(){
                System.out.println(c);
            }
            public add(){
                System.out.println("Test");
            }
    }

// Main class
class Test{
    public static void main(String args[]){
             A a=new A(1,2);
             B b=new B(3,4,5);
             a.show(); // 1 2
             b.show(); // 5
             b.add(); // Hello
             a = b; // Super class reference to Sub class.
             a.show(); // 3,4
             b.show(); // 5
             a.add(); // Error. 
    }
}

I am actually confused at this statement where it gives error.
a.add();
I am confused here. Why it is so? I can not understand, as "a" has been referenced to "b" and we all know the concept of referece. So, why "a" can not call add() here? I know, you will say it's not implemented in A, i know, but why?

DISCUSS PLEASE...

I just counted 20 compiler errors in that code. Are you sure everything works how you think it does? Fix them and create an SSCCE that demonstrates what you're talking about, and we'll go from there.

Like I said. We can help you when you provide an SSCCE. Talking in theory is fine, but it is a thousand times easier when we have some common working code to which we can refer. Plus that code contains some things that make me suspect you're working off incorrect assumptions, so creating an SSCCE will help clear them up. It shouldn't take more than a minute (it will take less time than it took to write these posts).

Okay. I still had to fix your method declarations, but good enough.

The reason it's saying that the add() method is not defined in type A is because, well, it isn't. And your 'a' variable is of type A, so the compiler has no idea what subclass might be stored in it. For example, you could have an instance of some class C that extends A and does not have the add() method stored in the 'a' variable.

Here's an SSCCE that demonstrates what I'm talking about a little more clearly:

 
public class Test
{
 
	public static void main(String args[]){
		Animal animal = new Animal();
		Cat cat = new Cat();
		animal.eat();
		cat.meow();
 
		animal = cat;
		animal.meow(); //error
 
		animal = new Dog();
		animal.meow(); //error, for more obvious reasons
 
	}
}
 
class Animal{
	public void eat(){
		System.out.println("Animal eating.");
	}
}
 
class Cat extends Animal{
	public void meow(){
		System.out.println("Meow");
	}
}
 
class Dog extends Animal{
	public void bark(){
		System.out.println("Bark");
	}
}

Originally Posted by Mr.777

Kevin, what you tried to do is, you just referenved superclass to the subclass say dog, but you tried to call the function from cat.
I am not talking about this.

Yes, you are. That's the point. The compiler has no way to know what you're storing in the superclass variable. It has no way to know what methods the subclasses are going to have. So even though you might know that a Cat can meow, unless you explicitly store (or cast) the reference as a Cat, you won't be able to use any Cat methods. This is because the compiler has no way to know it's a Cat or a Dog or a Fish or whatever. The same exact thing applies to your code.

Originally Posted by Mr.777

Let's say, there is a function called "color()" in class Dog but not in class Animal.
And i do this,

Animal animal=new Dog();
animal.color(); // This will create an error, right???

So, why will animal.color() will give error????

Because you're storing it as an Animal, so the compiler has no way to know what's in there. What if you had a List of Animals, each being a different subclass? The compiler can't know what a reference is storing, only what the reference claims to be. So since all you're claiming to be is an Animal, you can only use methods available to the Animal class. That's exactly what my example shows. Please go over it again carefully.

Originally Posted by Mr.777

as you can see that animal is now referenced as dog. That is the actual confusion. I know, that it will generate error. But i want to know, why? What is the true reason?

No, the reference is still an Animal. The instance may be a Dog, but the compiler can't know that.

Originally Posted by Mr.777

According to the reference concept, say

String a="Hi";
String b="Bye";
a=b;
System.out.println(a);
System.out.println(b);

Both print statements will generate the same result. It's reference, actually the object is given the address of the right side object.

This doesn't have much to do with your original question. In this case, both a and b are Strings. But what would happen if you did something like this?

public class Test
{
 
	public static void main(String args[]){
		String string1 = "test1";
		Object string2 = string1;
 
		System.out.println(string2.toUpperCase());
	}
}

You have a fundamental misunderstanding of how inheritance works with regards to references and instances. I suggest you read up on the basic tutorials and try more examples before you claim that my code is not what you're talking about.

Mr. 777,

You have a logical error in your post that

We know that a super class's object can refer to a sub class object

'a' is not an object, it is a reference to an object of type A.

You have classes A and B where B is a subclass of A.

In your code, you write:

A a=new A(1,2);
B b=new B(3,4,5);

Then you go on to write:

 a = b; // Super class reference to Sub class.
a.show(); // 3,4
b.show(); // 5
a.add(); // Error.

a = b; is valid because b is of type B which is a sublclass of A, so it is a type of A

but your statement a.add(); is a huge mistake because there is no add() method in class A. Class B is the only class with an add() method. What you have done with a = b; is you have hidden all the things that makes a B a B and you have forced it to conform to A's definition. You can cast 'a' as a B and achieve its add functionality that way, but as long as you are not explicitly typecasting your 'a' variable, you are stuck with the public methods of A, and this only works because B is a type of A.

Kevin's example, while it may not be as clear as you need it to be, is spot on. He uses class Animal (Class A) and has a subclass Cat (Class B).
He creates the references and an instance of Cat and assigns it to Animal. He then tries to make the Animal meow.... but not all animals meow. Even though you may be able to tell the difference between a Cat and an Animal, the Java compiler doesn't. That's why I talked about casting. Casting tells the compiler explicitly that the Animal it is looking at is a Cat.

Consider that you have another class C that extends A and it has a method subtract. You apply the same logic as with B.

a = c;
a.subtract();

Do you expect this to work? It won't because the subtract() method is not a part of class A. It's all about being type-safe. Java is a strongly-typed language, so you can't get away with things like calling another classes methods.
Type safety - Wikipedia, the free encyclopedia

I don't know much about c++, but I would like to correct one thing from the Java perspective:

Originally Posted by ashutoshtiwari

Very simply if CAR is a base class and Mercedes is a subclass of it. No doubt every feature of CAR is in MERCEDES.But every feature of Mercedes is not in CAR, like the ABS,EPS..So it can't be referred by a base class CAR!!

A Mercedes can be referenced as a Car, like so:

Car car = new Mercedes();

However, without casting, the car variable cannot access the methods in Mercedes (which I think is what you meant).

That might just be a case of me being pedantic with your terminology, but I didn't want the OP to become more confused.