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What is the Output of following Java Program?abstract class Demo{    public int a;    Demo()    {        a = 10;    }     abstract public void set(); }class Test extends Demo{     final public void get()    {        System.out.println("a = " + a);    }     public static void main(String[] args)    {        Test obj = new Test();        obj.get();    }}Select one:a.Compile Time Errorb.a=10c.Runtime Exception

Question

What is the Output of following Java Program?abstract class Demo{    public int a;    Demo()    {        a = 10;    }     abstract public void set(); }class Test extends Demo{     final public void get()    {        System.out.println("a = " + a);    }     public static void main(String[] args)    {        Test obj = new Test();        obj.get();    }}Select one:a.Compile Time Errorb.a=10c.Runtime Exception

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Solution

The output of the given Java program will be a Compile Time Error.

Here's why:

In the given Java program, the class 'Demo' is declared as abstract and it contains an abstract method 'set()'. However, when 'Demo' is extended by the class 'Test', the 'set()' method is not implemented or overridden.

In Java, when a regular (i.e., non-abstract) class extends an abstract class, it must provide implementations (i.e., bodies) for all of the abstract methods in the superclass. If it doesn't, the subclass itself must also be declared abstract.

Since 'Test' does not implement the abstract method 'set()' from 'Demo' and 'Test' is not declared as abstract, the Java compiler will produce a compile time error.

So, the correct answer is a. Compile Time Error.

This problem has been solved

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