java - 为什么异常类引用变量在java中打印消息？

Question

我是java新手，我不明白为什么异常类引用变量打印消息，而普通类的引用变量打印eclassname@jsjka为什么？

public class Exception11 {
    int x,y;

    public static void main(String[] args) {
        try{
        int total;
        Exception11 e=new Exception11();
        e.x=10;
        e.y=0;
        total=10/0;
        System.out.println("Value of refernce variable: "+e);
        System.out.println(total);
        } catch (ArithmeticException h) {
            System.out.println("number can not divide by the 0 Please try again");
            int total;
            Exception11 e=new Exception11();
            System.out.println("Value of refernce variable: "+e);
            System.out.println("Value of refernce variable: "+h);



        }

    }

}

回答-----------------------------------------

number can not divide by the 0 Please try again
Value of refernce variable: Exception11@4f1d0d
Value of refernce variable: java.lang.ArithmeticException: / by zero

Answer 1

您看到的是Object#toString代表你的类(class)。相反，ArithmeticException 已经覆盖了此方法。您需要在Exception11

中重写此方法

@Override
public String toString() {
    return "Exception11 [x=" + x + ", y=" + y + "]";
}