Java带给你的并不仅仅是面向对象、开放、平台无关、易用、安全和“Write once, run anywhere”等软件开发方面的优势，更重要的一点是，它提供了一种新颖的表达思想的方式，一种全新的思维模式。下面一起来看看Java2程序员考试题及答案!
例题1：
Choose the three valid identifiers from those listed below.
A. IDoLikeTheLongNameClass
B. $byte
C. const
D. _ok
E. 3_case
解答：A, B, D
点评：Java中的标示符必须是字母、美元符($)或下划线(_)开头。关键字与保留字不能作为标示符。选项C中的const是Java的保留字，所以不能作标示符。选项E中的3_case以数字开头，违反了Java的规则。
例题2：
How can you force garbage collection of an object?
A. Garbage collection cannot be forced
B. Call System.gc().
C. Call System.gc(), passing in a reference to the object to be garbage collected.
D. Call Runtime.gc().
E. Set all references to the object to new values(null, for example).
解答：A
点评：在Java中垃圾收集是不能被强迫立即执行的。调用System.gc()或Runtime.gc()静态方法不能保证垃圾收集器的立即执行，因为，也许存在着更高优先级的线程。所以选项B、D不正确。选项C的错误在于，System.gc()方法是不接受参数的。选项E中的方法可以使对象在下次垃圾收集器运行时被收集。
例题3：
Consider the following class:
1. class Test(int i) {
2. void test(int i) {
3. System.out.println(“I am an int.”);
4. }
5. void test(String s) {
6. System.out.println(“I am a string.”);
7. }
8.
9. public static void main(String args[]) {
10. Test t=new Test();
11. char ch=“y”;
12. t.test(ch);
13. }
14. }
Which of the statements below is true?(Choose one.)
A. Line 5 will not compile, because void methods cannot be overridden.
B. Line 12 will not compile, because there is no version of test() that rakes a char argument.
C. The code will compile but will throw an exception at line 12.
D. The code will compile and produce the following output: I am an int.
E. The code will compile and produce the following output: I am a String.
解答：D
点评：在第12行，16位长的char型变量ch在编译时会自动转化为一个32位长的int型，并在运行时传给void test(int i)方法。
例题4：
Which of the following lines of code will compile without error?
A.
int i=0;
if (i) {
System.out.println(“Hi”);
}
B.
boolean b=true;
boolean b2=true;
if(b==b2) {
System.out.println(“So true”);
}
C.
int i=1;
int j=2;
if(i==1|| j==2)
System.out.println(“OK”);
D.
int i=1;
int j=2;
if (i==1 &| j==2)
System.out.println(“OK”);
解答：B, C
点评：选项A错，因为if语句后需要一个boolean类型的表达式。逻辑操作有^、&、| 和 &&、||,但是“&|”是非法的，所以选项D不正确。
例题5：
Which two demonstrate a "has a" relationship? (Choose two)
A. public interface Person { }
public class Employee extends Person{ }
B. public interface Shape { }
public interface Rectandle extends Shape { }
C. public interface Colorable { }
public class Shape implements Colorable
{ }
D. public class Species{ }
public class Animal{private Species species;}
E. interface Component{ }
class Container implements Component{
private Component[] children;
}
解答：D, E
点评： 在Java中代码重用有两种可能的方式，即组合(“has a”关系)和继承(“is a”关系)。“has a”关系是通过定义类的属性的方式实现的;而“is a”关系是通过类继承实现的。本例中选项A、B、C体现了“is a”关系;选项D、E体现了“has a”关系。
例题6：
Which two statements are true for the class java.util.TreeSet? (Choose two)
A. The elements in the collection are ordered.
B. The collection is guaranteed to be immutable.
C. The elements in the collection are guaranteed to be unique.
D. The elements in the collection are accessed using a unique key.
E. The elements in the collection are guaranteed to be synchronized
解答：A, C
点评：TreeSet类实现了Set接口。Set的特点是其中的元素惟一，选项C正确。由于采用了树形存储方式，将元素有序地组织起来，所以选项A也正确。
例题7：
True or False: Readers have methods that can read and return floats and doubles.
A. Ture
B. False
解答：B
点评： Reader/Writer只处理Unicode字符的输入输出。float和double可以通过stream进行I/O.
例题8：
What does the following paint() method draw?
1. public void paint(Graphics g) {
2. g.drawString(“Any question”, 10, 0);
3. }
A. The string “Any question?”, with its top-left corner at 10,0
B. A little squiggle coming down from the top of the component.
解答：B
点评：drawString(String str, int x, int y)方法是使用当前的颜色和字符，将str的内容显示出来，并且最左的字符的基线从(x,y)开始。在本题中，y=0，所以基线位于最顶端。我们只能看到下行字母的一部分，即字母‘y’、‘q’的下半部分。
例题9：
What happens when you try to compile and run the following application? Choose all correct options.
1. public class Z {
2. public static void main(String[] args) {
3. new Z();
4. }
5.
6. Z() {
7. Z alias1 = this;
8. Z alias2 = this;
9. synchronized(alias1) {
10. try {
11. alias2.wait();
12. System.out.println(“DONE WAITING”);
13. }
14. catch (InterruptedException e) {
15. System.out.println(“INTERR
UPTED”);
16. }
17. catch (Exception e) {
18. System.out.println(“OTHER EXCEPTION”);
19. }
20. finally {
21. System.out.println
(“FINALLY”);
22. }
23. }
24. System.out.println(“ALL DONE”);
25. }
26. }
A. The application compiles but doesn't print anything.
B. The application compiles and print “DONE WAITING”
C. The application compiles and print “FINALLY”
D. The application compiles and print “ALL DONE”
E. The application compiles and print “INTERRUPTED”
解答：A
点评：在Java中，每一个对象都有锁。任何时候，该锁都至多由一个线程控制。由于alias1与alias2指向同一对象Z，在执行第11行前，线程拥有对象Z的锁。在执行完第11行以后，该线程释放了对象Z的锁，进入等待池。但此后没有线程调用对象Z的notify()和notifyAll()方法，所以该进程一直处于等待状态，没有输出。
例题10：
Which statement or statements are true about the code listed below? Choose three.
1. public class MyTextArea extends TextArea {
2. public MyTextArea(int nrows, int ncols) {
3. enableEvents(AWTEvent.TEXT_
EVENT_MASK);
4. }
5.
6. public void processTextEvent
(TextEvent te) {
7. System.out.println(“Processing a text event.”);
8. }
9. }
A. The source code must appear in a file called MyTextArea.java
B. Between lines 2 and 3, a call should be made to super(nrows, ncols) so that the new component will have the correct size.
C. At line 6, the return type of processTextEvent() should be declared boolean, not void.
D. Between lines 7 and 8, the following code should appear: return true.
E. Between lines 7 and 8, the following code should appear: super.processTextEvent(te).
解答：A, B, E
点评：由于类是public，所以文件名必须与之对应，选项A正确。如果不在2、3行之间加上super(nrows,ncols)的话，则会调用无参数构建器TextArea(), 使nrows、ncols信息丢失，故选项B正确。在Java2中，所有的事件处理方法都不返回值，选项C、D错误。选项E正确，因为如果不加super.processTextEvent(te)，注册的listener将不会被唤醒。
