多多问题，请大家帮忙。
在网上找了份题，以下我用（）括起来的是默认答案。
1》
public class Test{
public static void stringReplace(String text){
text=text.replace('j','l');
}
public static void bufferReplace(StringBuffer text){
text=text.append("c");
}
public static void main(String args[]){
String textString=new String("java");
StringBuffer textBuffer=new StringBuffer("java");
StringReplace(textString);
bufferReplace(textBuffer);
System.out.println(textString+textBuffer); javajavac
} string 类的对象本来就不可以改变，但StringBuffer 对象是可以改变的？？？
}
what is the output? （javajavac）
string 类的对象本来就不可以改变，但StringBuffer 对象是可以改变的？？？
不是说JAVA 的调用都是传值调用吗？何况上面两个函数都是VOID呀？
2》 public class ConstOver{
public ConstOver(int x, int y, int z){}
}
which two overload the ConstOver constructor?
（A.）ConstOver(){}
B.protected int ConstOver(){}
（C.）private ConstOver(int z, int y, byte x){}
D.public void ConstOver(byte x, byte y, byte z){} //这个为什么不可以？？
E.public Object ConstOver(int x, int y, int z){}
3》class BaseClass{
private float x=1.0f;
private float getVar(){return x;}
}
class SubClass extends BaseClass{
private float x=2.0f;
//insert code
}
what are true to override getVar()?
（A.）float getVar(){
B.public float getVar(){
C.public double getVar(){
D.protected float getVar(){ //why not?????
E.public float getVar(float f){
4>> class A implements Runnable{
int i;
public void run(){
try{
Thread.sleep(5000);
i=10;
}catch(InterruptException e){}
}
}
public static void main(String[] args){
try{
A a=new A();
Thread t=new Thread(a);
t.start();
17)
int j=a.i;
19)
}catch(Exception e){}
}
}
what be added at line line 17, ensure j=10 at line 19?
A. a.wait(); B. t.wait();
( C.) t.join();//不知道这个方法是干什么的？？？
D.t.yield();
E.t.notify(); F. a.notify(); G.t.interrupt();
5》 public class Test{
public static void main(String[] args){
StringBuffer a=new StringBufer("A");
StringBuffer b=new StringBufer("B");
operate(a,b);
System.out.pintln(a+","+b);
}
public static void operate(StringBuffer x, StringBuffer y){
x.append(y);
y=x; //y不可以指向x的对象吗？
}
}
what is the output? （ab,b） 和第一题很象
6》which interface Hashtable implements?
（ A.） java.util.Map //同意的请举手
B. java.util.List
C. java.util.Hashable
D. java.util.Collection
7》57. which two interfaces provide the capability to store objects using a key-value pair? //同上题。
（A.） java.util.Map
B. java.util.Set
C. java.util.List
D. java.util.SortedSet
（ E.） java.util.SortedMap
F. java.util.Collection
8》61. which two statements declare an array capable of 10 ints? //这里是说十个数还是十bit?
（A）. int[] foo;
（B）. int foo[]; //这里是说十个数还是十bit?
C. int foo[10];
D. Object[] foo;
E. Object foo[10];
9>>是不是抽象类的子类必须把自己实现的方法定义为public?
10》83. byte[] array1,array2[]//array2[]是个多维数组？？？
byte array3[][]
byte[][] array4
if each has been initialized, which statement will cause a compile error?
（A）array2=array1 B.array2=array3 C.array2=array4
10》 int i=1,j=10;
do{
if(i>j)continue;
j--;
}while(++i<6);
what are the vale of i and j?
（A）i=6,j=5 //我想i=6,j=4
B.i=5,j=5
C.i=6,j=4
D.i=5,j=6
E.i=6,j=6
11>> public class Foo{
public static void main(String[] args){
int i=1;
int j=i++;
if((i>++j)&&(i++==j)){ //i>++j这个已经是FALSE了，省下不执行。
i+=j;
}
}
}
what is the final value i and j? (i=2,j=2)
12>>public class X{
public static void main(String[] args){
int[] a=new int[1];
4} modify(a);
System.out.println(a[0]);
}
public static void modify(int[] a){ //这里是传地址还是传值调用
9} a[0]++;}
}
what is the result?
A.The program runs and prints "0";
（ B）.The program runs and prints "1";
C.The program runs but aborts with an exception;
13》class Super{
public int i=0;
public Super(String text){
i=1;
}
}
public class Sub extends Super{
public Sub(String text){
i=2;
}
public static void main(String args[]){
Sub sub=new Sub("Hello");
System.out.println(sub.i);
}
}
what is the result?
A. compile will fail //why
B. compile success and print "0"
C. compile success and print "1"
D. compile success and print "2" //Sub都承继了i,为什么不可以访问。
问题点数：100、回复次数：3Top
1 楼helpall（was jl）回复于 2003-02-04 00:23:05 得分 50
The difference between String and StringBuffer is that StringBuffer is mutable while String is not.
So String.replace(...) will create a new String. StringBuffer.repleace(...) will use the same StringBuffer object. Actually for the StringBuffer case, even if you use text.append("c"); The result will be the same.Top
2 楼helpall（was jl）回复于 2003-02-04 00:44:48 得分 50
2》 constructor can not have return type :-)
4》 wait for the t thread to die before continue processing.
5》 Well, inside the operate(..) function, y is indeed "ab" if you want to print the value. But remember this y is only a copy of the a, which is a pointer to the original "b". So y=x just let y points to "ab", then this y is discarded, the b outside of the function remains the same.
6》Why don't you agree with it? Hashtable can put key and value, it is a kind of Map.
12》 Address. Address does not change. Value inside changed.Top
3 楼orange2002（orange）回复于 2003-02-04 10:50:13 得分 0
多谢 helpall ,新年快乐，恭喜发财！Top
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