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Oracle

1Z0-809 · Question #48

Given: 1. abstract class Shape { 2. Shape ( ) { System.out.println ("Shape"); } 3. protected void area ( ) { System.out.println ("Shape"); } 4. } 5. 6. class Square extends Shape { 7. int side; 8…

The correct answer is D. At line 17, insert super (x); F. At line 20, use public void area () {. D is correct because Rectangle extends Square, and Square defines only a parameterized constructor Square(int side) - meaning Java does not auto-generate a no-arg constructor. Without an explicit super(x) call, Rectangle's constructor would try to invoke a non-existent…

Question

Given:
  1. abstract class Shape {
  2. Shape ( ) { System.out.println ("Shape"); }
  3. protected void area ( ) { System.out.println ("Shape"); }
  4. }
  6. class Square extends Shape {
  7. int side;
  8. Square (int side) {
  9. /* insert code here */
  10. this.side = side;
  11. }
  12. public void area ( ) { System.out.println ("Square"); }
  13. }
  14. class Rectangle extends Square {
  15. int len, br;
  16. Rectangle (int x, int y) {
  17. /* insert code here */
  18. len = x; br = y;
  19. }
  20. void area ( ) { System.out.println ("Rectangle"); }
  21. } Which two modifications enable the code to compile?

Options

  • AAt line 1, remove abstract
  • BAt line 9, insert super();
  • CAt line 12, remove public
  • DAt line 17, insert super (x);
  • EAt line 17, insert super (); super.side = x;
  • FAt line 20, use public void area () {

How the community answered

(44 responses)
  • A
    5% (2)
  • C
    2% (1)
  • D
    84% (37)
  • E
    9% (4)

Explanation

D is correct because Rectangle extends Square, and Square defines only a parameterized constructor Square(int side) - meaning Java does not auto-generate a no-arg constructor. Without an explicit super(x) call, Rectangle's constructor would try to invoke a non-existent Square(), causing a compile error. F is correct because Java prohibits narrowing a method's access modifier when overriding: Square.area() is public, so Rectangle.area() cannot silently drop to package-private (default) - it must be public or wider.

Why the distractors fail:

  • A - Removing abstract is unnecessary; an abstract class with a no-arg constructor is valid Java and isn't causing any error here.
  • B - Square's constructor already compiles without an explicit super() because Shape does have a no-arg constructor, so the compiler inserts the call automatically.
  • C - Removing public from Square.area() would create a new error: overriding Shape's protected method with package-private access is itself an illegal narrowing.
  • E - super() would invoke Square's no-arg constructor, which doesn't exist, so this fails to compile.

Memory tip: Think "two constructor rules": (1) if a parent has no default constructor, you must call super(args) explicitly; (2) when overriding, access can only go up (protected → public is fine; public → default is not). Both rules come from the hierarchy flowing downward.

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