What is the output of the following code? 1: abstract class Reptile { 2: public final void layEggs() { System.out.println("Reptile laying eggs"); } 3: public static void main(String[] args) { 4: Reptile reptile = new Lizard(); 5: reptile.layEggs(); 6: } 7: } 8: public class Lizard extends Reptile { 9: public void layEggs() { System.out.println("Lizard laying eggs"); } 10: } *1 pointA. Reptile laying eggsB. Lizard laying eggsC. The code will not compile because of line 4.D. The code will not compile because of line 5.E. The code will not compile because of line 9.
Question
What is the output of the following code? 1: abstract class Reptile { 2: public final void layEggs() { System.out.println("Reptile laying eggs"); } 3: public static void main(String[] args) { 4: Reptile reptile = new Lizard(); 5: reptile.layEggs(); 6: } 7: } 8: public class Lizard extends Reptile { 9: public void layEggs() { System.out.println("Lizard laying eggs"); } 10: } *1 pointA. Reptile laying eggsB. Lizard laying eggsC. The code will not compile because of line 4.D. The code will not compile because of line 5.E. The code will not compile because of line 9.
Solution
The correct answer is E. The code will not compile because of line 9.
Here's why:
In Java, the keyword 'final' when used with a method means that the method cannot be overridden in a subclass. In the given code, the method 'layEggs()' in the 'Reptile' class is declared as final. Therefore, when you try to override this method in the 'Lizard' class (which is a subclass of 'Reptile'), it will cause a compile-time error. Hence, the code will not compile because of line 9.
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