Core Java · interview question

Given the following code, why does it compile and execute successfully? interface Order { void process(); } class OnlineOrder implements Order { public void process() { System.out.println("Online Order Processed"); } } public class Main { public static void main(String[] args) { Order order = new OnlineOrder(); order.process(); } }?

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

java

interface Order { void process(); }
class OnlineOrder implements Order { public void process() { System.out.println("Online Order Processed"); } }
public class Main { public static void main(String[] args) { Order order = new OnlineOrder(); order.process(); } }

The process method is not overridden.Interfaces allow upcasting, enabling polymorphic behavior.The code uses abstract classes, not interfaces.Interfaces do not require method implementation.

Answer and explanation

Correct answer

B. Interfaces allow upcasting, enabling polymorphic behavior.

Why

Interfaces in Java allow upcasting, which means an OnlineOrder object can be treated as an Order reference. This enables polymorphic behavior, allowing the process method to be called on the interface reference, executing the overridden method in the OnlineOrder class.

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