Question

java 十进制转二进制！

class：ShowBinary
数字验证method：public int checkDigit(String data) throws MyWorngNumberException
二进制转换method：Public String printBits (int n)
这个是要求...在线等....哪位高人能帮个忙啊...急！

Answer 1

import javax.swing.JOptionPane;

public class ShowBinary {
public int checkDigit(String data) throws MyWorngNumberException{
try{
return Integer.valueOf(data).intValue();
}catch(Exception e){
e.printStackTrace();
throw new MyWorngNumberException(data);
}
}
public String printBits(int n){
StringBuffer result = new StringBuffer();
int temp = n;
while(n!=0){
temp =n%2;
n =n/2;
result.insert(0, temp);
}
return result.toString();
}
public static void main(String[] args){
int realvalue = Integer.MIN_VALUE ;
ShowBinary test = new ShowBinary();
String data = JOptionPane.showInputDialog("请输入一个整数", "10");
try{
realvalue = test.checkDigit(data);
}catch(MyWorngNumberException e){
JOptionPane.showMessageDialog(null, "输入的不是一个整数","警告", JOptionPane.ERROR_MESSAGE, null);
}
if(realvalue!=Integer.MIN_VALUE ){
JOptionPane.showMessageDialog(null, "转换结果："+test.printBits(realvalue),"提示", JOptionPane.INFORMATION_MESSAGE, null);
}
}
private class MyWorngNumberException extends Exception{

private static final long serialVersionUID = 9009817559695908225L;

public MyWorngNumberException (String data){
super(data+"is a worng number!");
}
}
}

Answer 2

楼上都写得那么复杂干嘛，其实很简单的

public class ShowBinary {

/**
* 数字验证
* @param data
* @return
* @throws MyWorngNumberException
*/
public static int checkDigit(String data) throws MyWorngNumberException {
try {
return Integer.parseInt(data);
} catch(Exception e) {
throw new MyWorngNumberException();
}
}

/**
* 二进制转换
* @param n
* @return 二进制
*/
public static String printBits (int n) {
return Integer.toBinaryString(n);
}

public static void main(String[] args) throws MyWorngNumberException {
String s = "3";
System.out.println(printBits(checkDigit(s)));
}
}

class MyWorngNumberException extends Exception {
}

Answer 3

检查数字
checkDigit(String data){
try{
return Integer.valueOf(data).intValue();
}catch(Exception ex){
throw new MyWorngNumberException();
}
}
很简单，就一个方法调用。
String binStr = Integer.toBinaryString(n);
System.out.println(binStr);

Answer 4

public class TenToTwo {

　　public static String change(long DecNumber) {

　　String BinNumber = "";

　　for (int i = 64; i >= 0; i--) {

　　if (DecNumber / (long) (Math.pow(2, i)) == 1) {

　　BinNumber = BinNumber + "1";

　　DecNumber = (long) (DecNumber - Math.pow(2, i));

　　} else {

　　if (!BinNumber.equals("")) {

　　BinNumber = BinNumber + "0";

　　}

　　}

　　}

　　return BinNumber;

　　}

　　/**

　　* @param args

　　*/

　　public static void main(String[] args) {

　　long DecNumber = 456;

　　System.out.println(TenToTwo.change(DecNumber));

　　}

　　}