如何在java中修复循环while/try catch错误

Question

我正在创建一个简单的计算器程序（这个 java 编程的第一周）。

问题背景：只有 5 个选项有效。 （1-加法；2-减法；3-乘法；4.除法；5.退出）。当用户输入 1-4 选项时，将填充结果，但用户需要循环返回以重新输入数据，直到选择了选项 5。 5是退出程序（结束循环/程序的唯一方法）。我的问题： 1. 如何阻止 try-catch 不停地运行？有没有更好的方法来实现 try-catch？例如，处理字符串数据错误消息。理想情况下，如果输入字符串，代码应该循环返回以通过生成消息“请重新输入数字...”再次重新启动，直到用户输入有效数字 2. 我正在尝试在主类中使用尽可能多的静态方法。我不确定我的做法是否准确？

这是我输入的代码：

    12 2 
    //-everything works well.
    2 //-enter again 
    s s (string entry-error) 

然后，将填充以下消息：

    "You have entered invalid floats. please re-enter:  
    Exception in thread "main" java.util.InputMismatchException
        ...
        at calculator.performSubtract(calculator.java:65)
        at calculator.main(calculator.java:34)" 

代码（示例）

   public class calculator {
//use static methods to implement the program 
    static Scanner userInput = new Scanner(System.in);
    static int userChoice;
    static float numberOne;
    static float numberTwo; 
    static float answer; 
    static int choice;

    public static void main(String[] args) {
       do {
       //this menu statement has to be repeated unless 5 is entered (exit the 
      //program) 
        System.out.println("Welcome to <dj's> Handy Calculator\n\n\t \1. Addition \n\t 2. Subtraction\n\t 3. Multiplication\n\t 4. Division\n\t 5. Exit\n\n");
        System.out.print("What would you like to do? ");

        try {   
        choice = userInput.nextInt();
        }catch (InputMismatchException e) {
            continue;
        }
        switch (choice) {
        case 2: performSubtract();
        break;
        ...
        case 5: exit();
        break;
        } 
        }while(choice >0 && choice <5);
        userInput.close();
    }

    public static void performSubtract () {
        //catch error statement. 
        try {
        numberOne = userInput.nextFloat();
        numberTwo= userInput.nextFloat();
        answer= numberOne-numberTwo;
        } catch (Exception e) {
        System.out.println("You have entered invalid floats. please re-enter:  ");
        numberOne = userInput.nextFloat();
        numberTwo= userInput.nextFloat();
        }
        System.out.printf("Please enter two floats to subtraction, separated by a space: %.1f %.1f\n", numberOne, numberTwo);
        System.out.printf("Result of subtraction %.1f and %.1f is %.1f\n", numberOne, numberOne, answer);
        System.out.println("\nPress enter key to continue...");
    }

}

Answer 1

我认为问题在于您没有从扫描仪中清除问题令牌。 您的 catch 语句会打印一条错误消息，然后再尝试将相同的标记解析为 int 或 float。

您可以在这里查看：https://www.geeksforgeeks.org/scanner-nextint-method-in-java-with-examples/

看来您需要调用 userInput.next() 才能越过无效令牌。

此外，如果您愿意，hasNextInt() 可以让您完全避免遇到问题。

Answer 2

您的错误在于Scanner.nextFloat 在读取无效输入时不会推进当前令牌。这意味着当您在 catch 语句中再次调用 nextFloat 两次时，您将再次读取标记 s 和 s，其中第一个将导致再次抛出 InputMismatchException。您应该将 performSubtract 方法更改为如下所示：

public static void performSubtract () {
    //catch errors
    System.out.println("Please enter two floats to subtraction, separated by a space");
    userInput.nextLine();//ignore the new line
    do {
        try {
            String[] nextLineTokens = userInput.nextLine().split("\\s+");
            if(nextLineTokens.length != 2)
            {
                System.out.println("You have not entered two floats. please re-enter:");
                continue;
            }
            numberOne = Float.parseFloat(nextLineTokens[0]);
            numberTwo = Float.parseFloat(nextLineTokens[1]);
            answer= numberOne-numberTwo;
            break;
        }
        catch (Exception e) {
            System.out.println("You have entered invalid floats. please re-enter:  ");
        }
    } while (true);
    System.out.printf("You entered: %.1f %.1f\n", numberOne, numberTwo);
    System.out.printf("Result of subtraction %.1f minus %.1f is %.1f\n", numberOne, numberTwo, answer);
    System.out.println("\nPress enter key to continue...");
    userInput.nextLine();
}

此外，如果您输入了无效的输入，您的解析代码将继续，但如果您输入的数字不是 1-5，则会退出。如果这是您第一次读入输入，则代码也会因无效输入而退出。你可能应该改变你的解析迭代循环：

public static void main(String[] args) {
    while(choice != 5) {
        //this menu statement has to be repeated unless 5 is entered (exit the 
        //program) 
        System.out.println("Welcome to <dj's> Handy Calculator\n\n\t 1. Addition \n\t 2. Subtraction\n\t 3. Multiplication\n\t 4. Division\n\t 5. Exit\n\n");
        System.out.print("What would you like to do? ");

        try {   
            choice = userInput.nextInt();
        }
        catch (InputMismatchException e) {
            userInput.next();
            continue;
        }
        switch (choice) {
            case 2: performSubtract();
            break;
            // ...
            case 5: exit();
            break;
        } 
    }
    userInput.close();
}

Answer 3

对于第一个问题：try-catch 块通常用于查看您的代码是否运行无误。通过你解释你想要做什么，我会在分配 numberOne 和 numberTwo 之前使用一个while循环，无论输入是否浮动：

// pseudo

while(input != float || input2 != float)
{
  print(please reenter)
}
numberOne = input
numberTwo = input2