Question

我尝试使用不同的选项（十六进制，二进制，八进制，十进制）制作桌面计算器，如果输入的值不是该模式，则会出现错误，例如＆＃34 ; BinaryInteger预计，用户输入909382＆＃34;并再次提示。现在唯一出现的错误是＆＃34;无效选项＆＃34;由于我的捕获（例外e）。如何为无效输入编写4个异常/错误。

import java.util.*;

public class IntDriver
{
   //declared fields
   private static LongInteger num1;
   private static LongInteger num2;
   private static String opr;
   private static int mode = 0;
   // Operators that will be used
   private final static String[] Operators = { "", "+", "-", "*", "/" };

   public static void main(String[] args)
   {
      Scanner kb = new Scanner(System.in);

      boolean run = true;
      while(run)
      {
         displayMenu();

         while(true)
         {
            try
            {
               System.out.print("Option or value --> ");
               String option = kb.nextLine();

               switch(option.toLowerCase())
               {
               //modes

                  // decimal mode
                  case "decimal":
                  case "dcm":
                  case "dec":
                     mode = 0;
                     break;

                  // binary mode
                  case "binary":
                  case "bin":
                     mode = 1;
                     break;

                  //octal mode
                  case "octal":
                  case "oct":
                     mode = 2;
                     break;

                  //hex mode
                  case "hexadecimal":
                  case "hex":
                     mode = 3;
                     break;

                  //quit
                  case "q":
                  case "quit":
                     System.out.println("Thank you. Have a nice day.");
                     run = false;
                     break;

               //Operators
                  //add
                  case "+":
                     opr = "+";
                     break;

                  //subtract   
                  case "-":
                     opr = "-";
                     break;

                  //multiply   
                  case "*":
                     opr = "*";
                     break;

                  //divide
                  case "/":
                     opr = "/";
                     break;

                  //equals
                  case "=":
                     operate();
                     break;

                  default:
                     LongInteger temp;
                     //with mode it's in:
                     switch (mode)
                     {
                        case 1:
                           temp = new BinaryInteger(option);
                           break;
                        case 2:
                           temp = new OctalInteger(option);
                           break;
                        case 3:
                           temp = new HexInteger(option);
                           break;
                        default:
                           temp = new DecInteger(option);
                           break;
                     }
                     // if no num1 = LongInteger temp.
                     if (num1 == null)
                     {
                        num1 = temp;
                     }
                     else
                     {
                        if (opr == null)
                        {
                           throw new UnsupportedOperationException();
                        }
                        num2 = temp;
                     }

                     break;
               }
               break;
            }
            catch (UnsupportedOperationException e)
            {
               System.out.println("Invalid option; operator not specified.\n");
            }
            // Invalid option entered.
            catch (Exception e)
            {
               System.out.println("Invalid option.\n");
            }
         }
      }
   }
   //Menu Display
   private static void displayMenu()
   {
      System.out.println();

      switch (mode)
      {
         case 1:
            System.out.println("Binary Mode");
            break;
         case 2:
            System.out.println("Octal Mode");
            break;
         case 3:
            System.out.println("Hexadecimal Mode");
            break;
         default:
            System.out.println("Decimal Mode");
            break;
      }

      if (num1 != null)
      {
         System.out.print(num1 + "\t");
      }

      if (opr != null)
      {
         System.out.print(opr + "\t");
      }

      if (num2 != null)
      {
         System.out.print(num2 + "\n");
      }
      System.out.println("\n\n\tModes:\t\t    Operators:");
      System.out.println("\tBin - Binary\t\t+");
      System.out.println("\tOct - Octal\t\t-");
      System.out.println("\tDcm - Decimal\t\t*");
      System.out.println("\tHex - Hexadecimal\t/");
      System.out.println("\tQ   - Quit\t\t=\n");
   }

   private static void operate() throws Exception
   {

      if (num1 == null || num2 == null || opr == null)
      {
         throw new Exception("Not enough numbers.");
      }

      switch (opr)
      {
         case "+":
            num1 = num1.calcValue(opr, num2, mode);
            break;
         case "-":
            num1 = num1.calcValue(opr, num2, mode);
            break;
         case "*":
            num1 = num1.calcValue(opr, num2, mode);
            break;
         case "/":
            num1 = num1.calcValue(opr, num2, mode);
            break;
         default:
            throw new Exception("Invalid operator.");
      }

      num2 = null;
      opr = "";
   }


}

Answer 1

你可以为每个案例延长java.lang.Exception次：

public class YourCustomException extends Exception {

  //copy paste and modify whatever constructors are relevant in your case.

}

..然后当你的模式输入无效时继续抛出这4个异常。不过，这在你的情况下是相当低效的。我认为你的案例很简单，你可以简单地做一些事情：

public class DecInteger {

    public DecInteger (String option)
    {
        if (isNotDecimal(option))
            throw new Exception("Input option " + option + " is not a decimal");
    }

}

..然后在你的catch区块你将需要做这样的事情：

catch(Exception ex) {
    System.out.println(ex.getMessage);
}

从技术上讲，抛出Exception类型的异常并不是一个好习惯，最好是创建更能描述实际异常的子类型，但是你的情况很简单，这并不是真的物质

尝试 - 捕获异常问题

1 个答案: