尝试在程序设计中捕获放置问题？

Question

我已经输入了 try catch，如图所示，程序仍然会跳到第二个和第三个输入问题，而不会在之后打印错误异常（“输入您的操作：加、减、除、乘或退出”）。此外，如果为第一个问题输入“退出”，则在程序完成之前第二个和第三个输入要求仍然循环，是否有任何方法可以在需要其他输入的情况下立即退出？ 欢迎任何建议

 import java.util.Scanner;
 import java.io.*;


 class Monday {
  public static void main(String[] args) {
    double n1,n2;
    boolean check = true;

    while(check) {
        System.out.println("Enter your operation: add, subtract, divide, multiply, or exit");
        Scanner myScan = new Scanner(System.in);
        String op = myScan.next();
        try {
            System.out.println("Enter your 1st number");
            try {
                n1 = myScan.nextDouble();
                System.out.println("Enter your 2nd number");
                n2 = myScan.nextDouble();
            } catch (Exception e) {
                System.out.println("This is my error");
                return;
            }

         /* System.out.println("Enter your 1st number");
            n1 = myScan.nextDouble();
            System.out.println("Enter your 2nd number");
            n2 = myScan.nextDouble();*/




            switch (op) {
                case"add":
                System.out.println("Your answer is "+ (n1 + n2));
                break;

                case"subtract":
                System.out.println("Your answer is "+ (n1 - n2));
                break;

                case"divide":
                System.out.println("Your answer is "+ (n1 / n2));
                break;

                case"multiply":
                System.out.println("Your answer is "+ (n1 * n2)) ;
                break;

                case"exit":
                System.out.println("Goodbye!");
                break;

            }


            if ("exit".equals(op))
            check = false;

        } catch (Exception e) {
            System.out.println("This is my error");

            System.exit(1);
        }
    }
}

}

Answer 1

你甚至不需要 try/catch。

double n1, n2;
System.out.println("Enter your 1st number");
if (myScan.hasNextDouble()) n1 = myScan.nextDouble();
else return;
System.out.println("Enter your 2nd number");
if (myScan.hasNextDouble()) n2 = myScan.nextDouble();
else return;

但如果你想使用它：

double n1, n2;
System.out.println("Enter your 1st number");
try {
    n1 = myScan.nextDouble();
    System.out.println("Enter your 2nd number");
    n2 = myScan.nextDouble();
} catch (Exception e) {
    System.out.println("This is my error");
    return;
}