java计算器（重新输入值）

我对交给我的java计算器作业有一些问题。我被告知要制作一个计算器，它可以执行非常基本的函数，捕捉异常，并允许您立即更正操作数或运算符的值（这就是我遇到的问题）。例如，这是控制台中应该发生的事情：

j * 6
Catch exception and print error message here and asking for new first operand
4
Answer: 4 * 6 = 24

或

到目前为止，我掌握了以下代码：

import java.util.*;
public class Calculator{

static int _state = 3;

public static void main(String[] args){

_state = 3;

System.out.println("Usage: operand1 operator operand2");
System.out.println(" (operands are integers)");
System.out.println(" (operators: + - * /");
@SuppressWarnings("resource")
Scanner in = new Scanner(System.in);
do{

try{
int result = 0;
int operand1 = 0;
int operand2 = 0;

String operator = "";
char op = ' ';

operand1 = in.nextInt();

operator = in.next();
op = operator.charAt(0);
operand2 = in.nextInt();

switch (op){

    default:
        System.out.println("You didn't insert a proper operator");
        break;
    case '+': result = operand1 + operand2;
        System.out.println("Answer: " + operand1 + ' ' + op + ' ' + operand2 + " = " + result );
        break;

    case '-': result = operand1 - operand2;
        System.out.println("Answer: " + operand1 + ' ' + op + ' ' + operand2 + " = " + result );
        break;

    case '*': result = operand1 * operand2;   
        System.out.println("Answer: " + operand1 + ' ' + op + ' ' + operand2 + " = " + result );
        break;

    case '/': result = operand1 / operand2;   
        System.out.println("Answer: " + operand1 + ' ' + op + ' ' + operand2 + " = " + result );
        break;
        }

}
    catch(ArithmeticException e){
       System.out.println("You can not divide by zero. Input a valid divider.");
    }
    catch (InputMismatchException e) {
      System.out.println("You must use proper numerals.");
  }   
} while(_state == 3);

}
}

---

在完成此操作之前，您需要做一些事情。我建议您通过扩展Exception类（或Exception类的子类）来创建自己的异常。您可以通过阅读以下内容来完成此操作：

现在假设您创建了MissingOperatorException，然后可以将其放入Switch/Case语句中

try{
    switch (op){

        default:
            System.out.println("You didn't insert a proper operator");
            throw MissingOperatorException;
            break;
        //some more code here
    }
}
//reach catch statements
catch(MissingOperatorException e){
   System.out.println("ERROR MESSAGE");
   //deal with case here by asking for new input
   //use scanner to get operand
   //you can do this by reusing code from above
   System.out.println("Please re-enter operand");
   operator = in.next();
   op = operator.charAt(0);
   //calculate answer again
   //print out answer
}

对于需要执行的每种类型的异常，都需要执行此操作。我看到三种可能的情况

失踪操作员

缺少左操作数

缺少右操作数

可能存在这些问题的组合，但是您编写的异常处理应该能够在处理过程中（一个接一个）处理这些问题

注意：您已经使用Scanner类来读取用户输入。我假设你了解扫描器的基本知识，但你的后续问题与此相矛盾。我真的会研究Java Scanner类。请查看以下内容：

---

用一个巨大的try块捕捉所有东西被认为是不好的做法。把试抓块放在你需要的地方

另外，人们并不真的喜欢为他们做别人的家庭作业。如果你有一个特定的问题，那很好，但是说“我必须为一个作业做这个，我怎么做所有的事情”可能不会成功。

你应该试试这个

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        boolean ret = true;
        while(ret){

        Scanner scn = new Scanner(System.in);
        String answer = "";
        String answer2 = "";
        System.out.println("Welcome to java calculator.....");
        System.out.print("\nEnter the first number: ");
        double numA = scn.nextDouble();
        System.out.print("Enter the second number: ");
        double numB = scn.nextDouble();
        System.out.println("What you want to calculate?");
        double result = 0;

        answer = scn.next();

        if(answer.equals("+")){
            result = numA + numB;
        }

        if(answer.equals("-")){
            result = numA - numB;
        }

        if(answer.equals("*")){
            result = numA * numB;
        }

        if(answer.equals("/")){
            result = numA / numB;
        }

        System.out.println("The result is: " + result);

        System.out.println("Do you want to continue?");
        answer2 = scn.next();

        if(answer2.equalsIgnoreCase("y")){
            ret = true;
            }

        else{
            System.out.println("Good bye....:)");
            System.exit(0);
            }
        }
    }
}

---

问题是什么…？我提供了这个控制台示例，您可以在其中立即重新输入以前无效的值，我需要有关如何编写计算器来实现这一点的想法/示例。在您说“//使用扫描仪获取操作数”的部分，我将如何做？这是几个星期来让我头晕目眩的部分。刚刚更新了我的帖子。你真的需要学习Scanner类是如何工作的。好吧，我想我已经掌握了重新输入值的部分。我会到处玩，看看能得到什么。谢谢你的帮助。只要你能从技术角度理解我的解决方案，你就可以走了。你仍然可以问问题。如果您有更详细的问题，我们会更容易帮助您。|=^）第一部分：说明。第二部分：110%的情况并非如此。

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        boolean ret = true;
        while(ret){

        Scanner scn = new Scanner(System.in);
        String answer = "";
        String answer2 = "";
        System.out.println("Welcome to java calculator.....");
        System.out.print("\nEnter the first number: ");
        double numA = scn.nextDouble();
        System.out.print("Enter the second number: ");
        double numB = scn.nextDouble();
        System.out.println("What you want to calculate?");
        double result = 0;

        answer = scn.next();

        if(answer.equals("+")){
            result = numA + numB;
        }

        if(answer.equals("-")){
            result = numA - numB;
        }

        if(answer.equals("*")){
            result = numA * numB;
        }

        if(answer.equals("/")){
            result = numA / numB;
        }

        System.out.println("The result is: " + result);

        System.out.println("Do you want to continue?");
        answer2 = scn.next();

        if(answer2.equalsIgnoreCase("y")){
            ret = true;
            }

        else{
            System.out.println("Good bye....:)");
            System.exit(0);
            }
        }
    }
}