Inheretance OCA 1Z0 -808

[Priyanka Leo]

Guys i am currently preparing for OCA 1Z0 -808 exam and i came across the below code which Outputs
Output :
Base
Dervied
Dervied


But what i expected was
Base
Base
Dervied

Could someone explain the logic of this. Thanks in advance guys!!

class Base
{
public void Print()
{System.out.println("Base");
}
}
class Derived extends Base
{
public void Print()
{
System.out.println("Dervied");
}
}
class mainn
{
public static void DoPrint (Base o)
{
o.Print();
}
public static void main(String[] args)
{
Base x = new Base();
Base y = new Derived();
Derived z = new Derived();
DoPrint(x);
DoPrint(y);
DoPrint(z);
}
}

[Mikalai Zaikin]

Hi Priyanka !

Welcome to Coderanch !

First off, let me repost your code in formatted way, so it will be more readable:

	class Base {
		public void Print() {
			System.out.println("Base");
		}
	}

	class Derived extends Base {
		public void Print() {
			System.out.println("Dervied");
		}
	}

	class mainn {
		public static void DoPrint(Base o) {
			o.Print();
		}

		public static void main(String[] args) {
			Base x = new Base();
			Base y = new Derived();
			Derived z = new Derived();
			DoPrint(x);
			DoPrint(y);
			DoPrint(z);
		}
	}

The behavior you see is called polymorphism, the method will always be called based on real instance of the object, not on the type of varible it referred by.

Some other hints: in Java by convention class names started with uppercase (mainn better name Mainn) and methods are started with lowercase (DoPrint better name doPrint, and Print better print)

[Mikalai Zaikin]

You studying for Java 8 exam, so this may be not in scope, but possibly it helps.

In Java 10 was introduced reserved type name var, so the main method could be rewritten as follows (again, won't work on Java 8):

	public static void main(String[] args) {
		var x = new Base();
		var y = new Derived();
		var z = new Derived();
		DoPrint(x);
		DoPrint(y);
		DoPrint(z);
	}

which also produces output:

Base
Dervied
Dervied

So, as you can see all 3 variables are declared identically, but point to different instances, and this is the instance of the class which decides what to print.

Hope it helps. 8-)

[Campbell Ritchie]

You will actually get slightly different semantics if you use var. In line 3, var will represent Derived, but in the Java8 version that was declared as Base. Of course the declared type won't affect how polymorphism works.

And, again, . . . . welcome to the Ranch

[Mikalai Zaikin]

Campbell Ritchie wrote:You will actually get slightly different semantics if you use var. In line 3, var will represent Derived, but in the Java8 version that was declared as Base. Of course the declared type won't affect how polymorphism works.

Yes, fully agree, as the type of variable in case of var will be derived from right hand side assignment type.

Then maybe another code demonstrating polymorphism would be :

	public static void main(String[] args) {
		Base x = new Base();
		Base y = new Derived();
		Base z = new Derived();
		DoPrint(x);
		DoPrint(y);
		DoPrint(z);
	}

resulting the same output:

Base
Dervied
Dervied

[Priyanka Leo]

thank you guys now i understood the logic!!