Given the code fragments:
and
What is the result?
Exam 1z0-809 All QuestionsBrowse all questions from this exam
Question 155
andCorrect Answer: B
The code will result in a compilation error because in Java, when you override a method in a subclass, the overridden method cannot declare a broader exception than the method in the superclass. In this case, the superclass method in the Video class declares an IOException, while the subclass method in the Game class declares a broader Exception, which is not allowed. Therefore, the code will not compile.
Discussion
ChirajOption: B
Rule: If the superclass method declares an exception, the subclass overridden method can declare the same, subclass exception or no exception but cannot declare parent exception.
jduarteOption: B
Answer B. Tested. Exception Exception is not compatible with throws clause in test.printValues()
Abdullah_RahahleahOption: B
Answer is B, tested class test{ public void printValues() throws IOException { System.out.println("Hi"); } } class test2 extends test{ public void printValues() throws Exception{ super.printValues(); System.out.println("hi2"); } }
steefaandOption: B
B is true because when overriding method you must use that same Exception class or its child.
iSnoverOption: B
Answer is B, tested class test{ public void printValues() throws IOException { System.out.println("Hi"); } } class test2 extends test{ public void printValues() throws Exception{ super.printValues(); System.out.println("hi2"); } }
WilsonKKerllOption: B
Answer is B.
mevltOption: A
The answer is A(tested) public class Video{ public void play() throws IOException{ System.out.print("Vidoe played."); } } class Game extends Video{ public void play() throws IOException{ super.play(); System.out.print("Game played."); } public static void main(String[] args) { try { new Game().play(); }catch (Exception e){ System.out.println(e.getClass()); } } }
mevlt
Nevermind.
Svetleto13Option: B
Answer is compilation error.Both classes shoult throw Exception to compile and get answer A.