SCJP 1.6版考題 080

3. interface Fish{}
4. class Perch implements Fish{}
5. class Walleye extends Perch{}
6. class Bluegill{}
7. public class Fisherman {
8.   public static void main(String[] args) {
9.     Fish f = new Walleye();
10.   Walleye w = new Walleye()
11.   Bluegill b = new Bluegill();
12.   if(f instanceof Perch) System.out.print("f-p ");
13.   if(w instanceof Fish) System.out.print("w-f ");
14.   if(b instanceof Fish) System.out.print("b-f ");
15. }
16.}


What is the result?
  A. w-f
  B. f-p w-f
  C. w-f b-f
  D. f-p w-f b-f
  E. Compilation fails.
  F. An exception is thrown at runtime.

解答

---

Ans: B

解說:

f instanceof Perch是判斷f所指向的物件型態是否為Perch, 
   f是指向Walleye物件，Walleye繼承Perch，所以12行if為true
w instanceof Fish是true，w是指向Walleye物件，
   Walleye繼承Perch，Perch實作Fish，所以Walleye也是Fish的型態
Bluegill類別與Fish沒有任何繼承關聯，所以b instanceof Fish為false