Ikk：新頁面: 1. class TestA{ 2. public void start(){System.out.println("TestA");} 3. } 4. public class TestB extends TestA{ 5. public void start(){System.out.println("TestB");} 6. publi...

2013-03-30T06:17:47Z

新頁面: 1. class TestA{ 2. public void start(){System.out.println("TestA");} 3. } 4. public class TestB extends TestA{ 5. public void start(){System.out.println("TestB");} 6. publi...

新頁面

1. class TestA{
2. public void start(){System.out.println("TestA");}
3. }
4. public class TestB extends TestA{
5. public void start(){System.out.println("TestB");}
6. public static void main(String[] args){
7. ((TestA)new TestB()).start();
8. }
9. }

What is the result?
A. TestA
B. TestB
C. Compilation fails.
D. An exception is thrown at runtime.

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<span style="font-size:medium;">解答</span>

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<span style="font-size: medium;">Ans: B </span>

<span style="font-size:medium;"> 解說:
new TestB()是產生TestB的物件，所以start方法為TestB的版本，答案是B

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{{SCJP 1.6版考題講解}}

SCJP 1.6版考題 076 - 修訂歷史

Ikk：新頁面: 1. class TestA{ 2. public void start(){System.out.println("TestA");} 3. } 4. public class TestB extends TestA{ 5. public void start(){System.out.println("TestB");} 6. publi...