Core Java · interview question

Why might the following code result in a compilation error? abstract class Service { abstract void execute(); } class Shipping extends Service { void execute() { System.out.println("Shipping executed"); } } public class Main { public static void main(String[] args) { Service s = new Shipping(); s.execute(); } }?

Multiple choiceHardchapter-5-curatedpremiumcore-javaupcasting-and-dynamic-dispatch

java

abstract class Service { abstract void execute(); }
class Shipping extends Service { void execute() { System.out.println("Shipping executed"); } }
public class Main { public static void main(String[] args) { Service s = new Shipping(); s.execute(); } }

Service cannot be instantiated directly.The execute method in Shipping is missing the override annotation.The execute method in Shipping must be static.The execute method in Shipping must be declared as public.

Answer and explanation

Correct answer

D. The execute method in Shipping must be declared as public.

Why

In Java, when a method is overridden, it must at least have the same access level as the method in the superclass. If Service's execute was public, Shipping's execute must also be public, otherwise, it results in a compilation error.

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