日期:2014-05-20  浏览次数:20819 次

一个小问题
下面是一段我自己设计的代码,目的是为了了解java.lang.Byte
Java code

public class Tbyte{
    public static void main(String args[]){
        Byte b1=new Byte("a");
        
        int i1;
        i1=b1.intValue();
        
        System.out.println(i1);
    }
}


虽然编译通过,但运行结果出现了下面的文字
--------------------Configuration: <Default>--------------------
Exception in thread "main" java.lang.NumberFormatException: For input string: "a"
  at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
  at java.lang.Integer.parseInt(Integer.java:449)
  at java.lang.Byte.parseByte(Byte.java:151)
  at java.lang.Byte.<init>(Byte.java:325)
  at Tbyte.main(Tbyte.java:3)

Process completed.

我不知道为什么会这样,求解释啊........

------解决方案--------------------
Java code
public class test {
    public static void main(String args[]) {
        Byte b1 = new Byte((byte)'a');

        int i1;
        i1 = b1.intValue();

        System.out.println(i1);
    }
}

------解决方案--------------------
Java code

String str = "127";//取值范围:"-128","-127",...."-2","-1","0","1","2",....,"126","127"
Byte b1 = new Byte(str);//因为它只有八个二进制位,十进制形式为大于-128小于127
int i1;
i1 = b1.intValue();

/*
Constructs a newly allocated Byte object that represents the byte value indicated by the String parameter. The string is converted to a byte value in exactly the manner used by the parseByte method for radix 10. 
*/