Java帶給你的並不僅僅是面向物件、開放、平臺無關、易用、安全和“Write once, run anywhere”等軟體開發方面的優勢，更重要的一點是，它提供了一種新穎的表達思想的方式，一種全新的思維模式。下面一起來看看Java2程式設計師考試題及答案!
例題1：
Choose the three valid identifiers from those listed below.
A. IDoLikeTheLongNameClass
B. $byte
C. const
D. _ok
E. 3_case
解答：A, B, D
點評：Java中的標示符必須是字母、美元符($)或下劃線(_)開頭。關鍵字與保留字不能作為標示符。選項C中的const是Java的保留字，所以不能作標示符。選項E中的3_case以數字開頭，違反了Java的規則。
例題2：
How can you force garbage collection of an object?
A. Garbage collection cannot be forced
B. Call ().
C. Call (), passing in a reference to the object to be garbage collected.
D. Call ().
E. Set all references to the object to new values(null, for example).
解答：A
點評：在Java中垃圾收集是不能被強迫立即執行的。呼叫()或()靜態方法不能保證垃圾收集器的立即執行，因為，也許存在著更高優先順序的執行緒。所以選項B、D不正確。選項C的錯誤在於，()方法是不接受引數的。選項E中的方法可以使物件在下次垃圾收集器執行時被收集。
例題3：
Consider the following class:
1. class Test(int i) {
2. void test(int i) {
3. tln(“I am an int.”);
4. }
5. void test(String s) {
6. tln(“I am a string.”);
7. }
8.
9. public static void main(String args[]) {
10. Test t=new Test();
11. char ch=“y”;
12. (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
點評：在第12行，16位長的char型變數ch在編譯時會自動轉化為一個32位長的int型，並在執行時傳給void test(int i)方法。
例題4：
Which of the following lines of code will compile without error?
A.
int i=0;
if (i) {
tln(“Hi”);
}
B.
boolean b=true;
boolean b2=true;
if(b==b2) {
tln(“So true”);
}
C.
int i=1;
int j=2;
if(i==1|| j==2)
tln(“OK”);
D.
int i=1;
int j=2;
if (i==1 &| j==2)
tln(“OK”);
解答：B, C
點評：選項A錯，因為if語句後需要一個boolean型別的表示式。邏輯操作有^、&、| 和 &&、||,但是“&|”是非法的，所以選項D不正確。
例題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
點評： 在Java中程式碼重用有兩種可能的`方式，即組合(“has a”關係)和繼承(“is a”關係)。“has a”關係是通過定義類的屬性的方式實現的;而“is a”關係是通過類繼承實現的。本例中選項A、B、C體現了“is a”關係;選項D、E體現了“has a”關係。
例題6：
Which two statements are true for the class Set? (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
點評：TreeSet類實現了Set介面。Set的特點是其中的元素惟一，選項C正確。由於採用了樹形儲存方式，將元素有序地組織起來，所以選項A也正確。
例題7：
True or False: Readers have methods that can read and return floats and doubles.
A. Ture
B. False
解答：B
點評： Reader/Writer只處理Unicode字元的輸入輸出。float和double可以通過stream進行I/O.
例題8：
What does the following paint() method draw?
1. public void paint(Graphics g) {
2. String(“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
點評：drawString(String str, int x, int y)方法是使用當前的顏色和字元，將str的內容顯示出來，並且最左的字元的基線從(x,y)開始。在本題中，y=0，所以基線位於最頂端。我們只能看到下行字母的一部分，即字母‘y’、‘q’的下半部分。
例題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. ();
12. tln(“DONE WAITING”);
13. }
14. catch (InterruptedException e) {
15. tln(“INTERR
UPTED”);
16. }
17. catch (Exception e) {
18. tln(“OTHER EXCEPTION”);
19. }
20. finally {
21. tln
(“FINALLY”);
22. }
23. }
24. tln(“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
點評：在Java中，每一個物件都有鎖。任何時候，該鎖都至多由一個執行緒控制。由於alias1與alias2指向同一物件Z，在執行第11行前，執行緒擁有物件Z的鎖。在執行完第11行以後，該執行緒釋放了物件Z的鎖，進入等待池。但此後沒有執行緒呼叫物件Z的notify()和notifyAll()方法，所以該程序一直處於等待狀態，沒有輸出。
例題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(_
EVENT_MASK);
4. }
5.
6. public void processTextEvent
(TextEvent te) {
7. tln(“Processing a text event.”);
8. }
9. }
A. The source code must appear in a file called
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: essTextEvent(te).
解答：A, B, E
點評：由於類是public，所以檔名必須與之對應，選項A正確。如果不在2、3行之間加上super(nrows,ncols)的話，則會呼叫無引數構建器TextArea(), 使nrows、ncols資訊丟失，故選項B正確。在Java2中，所有的事件處理方法都不返回值，選項C、D錯誤。選項E正確，因為如果不加essTextEvent(te)，註冊的listener將不會被喚醒。
