1. What gets printed when the following program
is compiled and run?
class Test {
public static void main(String args[]) {
int i;
do {
i++;
} while (i < 0);
System.out.println(i);
}
}
Select 1 correct answer:
A. The program does not compile as i is not initialized.
B. The program compiles but does not run.
C. The program compiles and runs but does not print anything.
D. The program prints 0.
E. The program prints 1.
答案：A：如果没有初始化便使用基本变量类型，会导致编译时异常，程序不能编译。
2. What gets printed when the following program
is compiled and run?
public class XYZ {
public static void main(String args[]) {
int i,j,k;
for (i = 0; i < 3; i++)
{
for(j=1; j < 4; j++)
{
for(k=2; k<5; k++)
{
if((i == j) && (j==k))
System.out.println(i);
} } } } }
Select 1 correct answer:
A. 0
B. 1
C. 2
D. 3
E. 4
答案：C
3. Given the following code :
class Base{}
public class MyCast extends Base{
static boolean b1=false;
static int i = -1;
static double d = 10.1;
public static void main(String argv[]){
MyCast m = new MyCast();
Base b = new Base();
//Here
}
}
Which of the following, if inserted at the comment //Here
will allow the code to compile and run without error?
Select 2 correct answers:
A. b = m;
B. m = b;
C. d = i;
D. b1 = i;
解析：A 从子类型到父类型的转换是扩展引用转换，不需要在运行时采取特殊的动作，不会在运行时抛出异常。
B 从超类型到子类型的转换是收缩引用转换，需要在运行时执行测试，以查明实际的引用值是否是新类型的合法值.如果不是， 则会抛出ClassCascException 。在这里，b本身不是MyCast类型的值，因此会导致运行时异常。
C 从int到double的转换是扩展基本转换，基本类型之间的扩展转换永远不会导致运行时异常。但从int或long到float，或者是从long到double都可能导致精度丢失。
D 不允许进行int和boolean之间的类型转换
答案：A、C
4. Given the following classes which of the following
will compile without error?
interface IFace{}
class CFace implements IFace{}
class Base{}
public class ObRef extends Base{
public static void main(String argv[])
{
ObRef ob = new ObRef();
Base b = new Base();
Object o1 = new Object();
IFace o2 = new CFace();
}
}
Select 3 correct answers:
A. o1 = o2;
B. b = ob;
C. ob = b;
D. o1 = b;
解析：A 任何对象都可以赋给Object类型的对象，正确
B 子类赋给超类是扩展引用转换，可以进行
C 收缩引用转换，错误
D 扩展引用转换，可以进行
答案 A B D
5. What is the result of compiling and running the following code?
1 public class Test {
2 static int total = 10;
3 public static void main (String args []) {
4 new Test();
5 }
6 public Test () {
7 System.out.println("In test");
8 System.out.println(this);
9 int temp = this.total;
10 if (temp > 5) {
11 System.out.println(temp);
12 }
13 }
14 }
Select 1 correct answer:
A. The class will not compile
B. The compiler reports an error at line 2
C. The compiler reports an error at line 9
D. The value 10 is one of the printed elements
E. The class compiles but generates a runtime error
答案：D 由于test类没有override toString方法，故在调用println(this)时会直接显示类的有关信息。
6. Carefully examine the following code:
public class StaticTest {
static { System.out.println("Hi there"); }
public void print() {
System.out.println("Hello");
}
public static void main(String args []) {
StaticTest st1 = new StaticTest();
st1.print();
StaticTest st2 = new StaticTest();
st2.print();
}}
When will the string "Hi there" be printed?
Select 1 correct answer:
A. Never
B. Each time a new instance is created
C. Once, when the class is first loaded into the JVM
D. Only when the static method is called explicitly
答案 C
Java虚拟机规范约定，类的初始化发生在首次主动使用时。静态变量和静态初始化块的先后，实际上取决于它们在类中出现的先后顺序。
7. Given the following code what is the effect of a being 5?
public class Test {
public void add(int a) {
loop: for (int i = 1; i < 3; i++){
for (int j = 1; j < 3; j++) {
if (a == 5) {
break loop;
}
System.out.println(i * j);
}
}
}
}
Select 1 correct answer:
A. Generates a runtime error
B. Throws an ArrayIndexOutOfBoundsException
C. Prints the values: 1, 2, 2, 4
D. Produces no output
解析：考察标记和break的使用。单纯使用break只跳出当前循环(一层)。如果在要跳出的循环前加标记，那么在加过标记的循环中的任何地方都可以使用break 标记 来跳出该循环
答案：D
8. Given the following class definition:
public class Droitwich{
class one{
private class two{
public void main(){
System.out.println("two");
}
}
}
}
Which of the following statements are true?
Select 1 correct answer:
A. The code will not compile because the classes are nested to
more than one level
B. The code will not compile because class two is marked as private
C. The code will compile and output the string two at runtime
D. The code will compile without error
答案：D
9. Given the following code:
class Base { static int oak=99; }
public class Doverdale extends Base{
public static void main(String argv[]){
Doverdale d = new Doverdale();
d.amethod();
}
public void amethod(){
//Here
}
}
Which of the following if placed after the comment //Here,
will compile and modify the value of the variable oak?
Select all possible answers:
A. super.oak=1;
B. oak=33;
C. Base.oak=22;
D. oak=50.1;
解析：当子类没有重新定义静态属性则子类的静态属性与父类的静态属性为同一个变量
当子类重新定义了父类静态属性则子类的静态属性与父类的静态属性是两个不同的变量
静态方法调用的是定义这个静态方法的类的静态属性。
答案：A B C
10. What will happen when you attempt to compile
and run the following code?
public class Inc {
public static void main(String argv[]) {
Inc inc = new Inc();
int i =0;
inc.fermin(i);
i = i++;
System.out.println(i);
}
void fermin(int i){
i++;
}
}
Select 1 correct answer:
A. Compile time error
B. Output of 2
C. Output of 1
D. Output of 0
答案：D
i++和++i是什么意思?_百度知道
答 都是i=i+1的意思，区别在于i++是i先不自加，在语句完后自加，++i先自加
11. In the code fragment below, what is the value of k
after line 3 executes?
1 int i = -1;
2 int j = i >> 3;
3 int k = j & 129;
A. -1
B. 0
C. 129
D. A very large negative number
E. A very large positive number
解析：本题考察了整数在java虚拟机中的二进制表示。
计算机中的符号数有三种表示方法，即原码、反码和补码。三种表示方法均有符号位和数值位两部分，符号位都是用0表示“正”，用1表示“负”，而数值位，三种表示方法各不相同。
在计算机系统中，数值一律用补码来表示和存储。原因在于，使用补码，可以将符号位和数值域统一处理;同时，加法和减法也可以统一处理。此外，补码与原码相互转换，其运算过程是相同的，不需要额外的硬件电路。
正整数的补码是其二进制表示，与原码相同，
求负整数的补码，将其对应正数二进制表示所有位取反(包括符号位，0变1，1变0)后加1。
简单起见，我们不表示出所有32位，仅表示十二位。
i : -1 1：:0000 0000 0001 -1: 1111 1111 11111
第二行，对i进行了移位操作，>>为带符号右移，其移出的空位需要用符号位补齐。格式为 需要移位的数字>>要移的位数
故j = 1111 1111 1111
第三行，&为按位与运算符，129=0000 1000 0001，k=0000 1000 0001,即129
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