What will be the output of the program? class Base
{
Base()
{
System.out.print("Base");
}
}
public class Alpha extends Base
{
public static void main(String[] args)
{
new Alpha(); /* Line 12 */
new Bas…

Question

What will be the output of the program? class Base
{ 
    Base()
    {
        System.out.print("Base");
    }
} 
public class Alpha extends Base
{ 
    public static void main(String[] args)
    { 
        new Alpha(); /* Line 12 */
        new Bas…

Accepted Answer

Option B is correct. It would be correct if the code had compiled, and the subclass Alpha had been saved in its own file. In this case Java supplies an implicit call from the sub-class constructor to the no-args constructor of the super-class theref…

What will be the output of the program? class Base { Base() { System.out.print("Base"); } } public class Alpha extends Base { public static void main(String[] args) { new Alpha(); /* Line 12 / new Base(); / Line 13 */ } }

More Questions from Declarations and Access Control

Discussion & Comments

What will be the output of the program? class Base { Base() { System.out.print("Base"); } } public class Alpha extends Base { public static void main(String[] args) { new Alpha(); /* Line 12 */ new Base(); /* Line 13 */ } }

More Questions from Declarations and Access Control

Discussion & Comments

What will be the output of the program? class Base { Base() { System.out.print("Base"); } } public class Alpha extends Base { public static void main(String[] args) { new Alpha(); /* Line 12 / new Base(); / Line 13 */ } }