java IO 获取输入流的问题

如下代码：
我在后台输入 1234567890

但是下面这段程序输入是：
1234567890

34567890

多了一段  34567890

 import java.io.*;  
public class Person {  
    public static void main(String args[]) {
        int b =10;
        byte[] x = new byte[b];
        try {
            System.out.println("please Input:");
            while ((b = System.in.read(x, 0, b)) != -1 && (b>0)) {
                String s = new String(x);
                System.out.println(s);
            }
        } catch (IOException e) {
            System.out.println(e.toString());
        }
    }  
} 

以下是我的理解：

    public static void main(String[] args){
        int b =10;
        byte[] x = new byte[b];
        try {
            System.out.println("please Input:");
/*            while ((b = System.in.read(x, 0, b)) != -1 && (b>0)) {
                String s = new String(x);
                System.out.println(s);
            }*/

            //下面一行运行完后 s=1234567890 b=10 x=[49, 50, 51, 52, 53, 54, 55, 56, 57, 48] 对应0-9的字节值
            b = System.in.read(x, 0, b);
            String s = new String(x);
            System.out.println(s);

            //下面一行运行完后x=[13, 10, 51, 52, 53, 54, 55, 56, 57, 48]因为刚才输入完0-9后又输入了回车符，对应输入了回车符和换行符，字节值为13和10
            b = System.in.read(x, 0, b);
            String s1 = new String(x);
            System.out.println(s1);

            //因为没有输入了，所以下面不会被运行
            b = System.in.read(x, 0, b);
            String s2 = new String(x);
            System.out.println(s2);

        } catch (IOException e) {
            System.out.println(e.toString());
        }
    }

关于System.in.read方法的说明可参考http://blog.csdn.net/foart/article/details/8211327