请告诉我该异常的描述未出现

问题描述

|

package javaapplication16;

class ExceptionoftheGods extends Exception {

    double b;

    ExceptionoftheGods (String msg){
    }

}

class mak {

    static void compute(int a) throws ExceptionoftheGods {
        System.out.println(\"Called Compute(\" + a + \")\");
        if(a > 7) {
            throw new ExceptionoftheGods(\"Dog\");
        }
        System.out.println(\"Normal Exit\");
    }    


    public static void main(String[] args) {
        try {
            compute(1);
            compute(9);
        } catch(ExceptionoftheGods e) {
            System.out.println(\" Caught \" + e);
        }
    }

}

输出值

run:
Called Compute(1)
Normal Exit
Called Compute(9)
 Caught javaapplication16.ExceptionoftheGods
BUILD SUCCESSFUL (total time: 0 seconds)

    

解决方法

        

ExceptionoftheGods(String msg) {
    super(msg); // missing
}

    ,        您需要在子类的构造函数中调用基本异常构造函数，否则实际上不会设置该消息：

ExceptionoftheGods(String msg) {
    super(msg);
}

另外，您正在打印异常对象本身，而不是异常消息。如果只想打印消息，则需要直接致电

getMessage()

，否则您将同时获得异常的类型和描述。 更改

System.out.println(\" Caught \" + e);

至

System.out.println(\" Caught \" + e.getMessage());

    ,        定义新方法时，必须定义其所有功能。如果要使其用作父类方法，则必须调用

super

您要么不定义构造函数（只保留默认的-empty-构造函数可用），要么修改构造函数。

ExceptionoftheGods (String msg){
  super(msg);
}