java - 如何实现猜谜游戏的自定义异常

Question

我必须创建一个最多允许 5 轮的猜谜游戏，并且用户输入必须在 1 到 10 之间(包含 1 和 10)。如果不满足这些条件，则可能会引发两个自定义异常(BadGuessException 和 TooManyGuessesException)。我陷入了如何进行异常处理的困境，因为我不确定如何让程序知道是否抛出和捕获这些自定义异常。

我为自定义异常创建了两个类:

public class BadGuessException extends Exception
{
  /**
   * no-arg constructor
   */
  public BadGuessException()
  {
    super("Sorry, that was an invalid guess!");
  }

  /**
   * parametrized constructor
   * @param message String message passed to super class's constructor
   */
  public BadGuessException(String message)
  {
    super(message);
  }
}

public class TooManyGuessesException extends Exception
{
  /**
   * no-arg constructor
   */
  public TooManyGuessesException()
  {
    super("Sorry, too many guesses!");
  }

  /**
   * parametrized constructor
   * @param guess integer value representing amount of guesses (turns)
   */
  public TooManyGuessesException(int guess)
  {
    super("Sorry, you guessed " + guess + " times!");
  }

}

在下面的代码中，我试图在抛出 TooManyGuessesException 之前允许最多五轮，并且我试图处理输入小于 1 和大于 10 的数字的异常。我需要只有一个 try-catch block (以及 NumberFormatException 的额外 catch 子句)。

import java.util.Random;
import java.util.*;

public class GuessingGame
{
  public static void main(String[] args)
  {
    //Scanner object to receive user input
    Scanner keyboard = new Scanner(System.in);

    //Create Random class object & random variable
    Random rng = new Random();
    int n = rng.nextInt(10 - 1 + 1) + 1;

    //Create incrementor for guessing turns
    int turn = 1;
    //Create variable for user input (guess)
    int guess;


    try
    {
      while(guess != n && turn <= 5)
      System.out.println("Guess a number between 1 and 10 inclusive.");
      System.out.println("Hint: the answer is " + n);
      guess = keyboard.nextInt();
      turn++;
      if(guess == n)
      {
        System.out.println("YOU WIN!\nIt took you " + turn + " attempts.");
      }

    }
    catch(BadGuessException e | TooManyGuessesException e)
    {
      if(guess < 1 || guess > 10)
        e.BadGuessException();
      if(turn > 5)
        e.TooManyGuessesException();
    }
    catch(NumberFormatException e)
    {
      System.out.println("Sorry, you entered an invalid number format.");
    }
  }
}

Answer 1

@Jack Flamp 刚才说的是完全正确的，但你不想检查猜测是否在 1 到 10 之间(含)，而是检查用户猜测是否与你的随机数匹配。

if (guess == n) {
    System.out.println("YOU WIN!\nIt took you " + turn + " attempts.");
} else {
    throw new BadGuessException();
}

如果您希望在turn变量超过5时抛出异常，请将其从while条件中删除，并在while范围的开头创建一个if:

if (turn > 5)
    throw new TooManyGuessesException();

PS:在 while 范围内加上括号，否则您将立即遇到 TooManyGuessesException

PS2:我想评论 Jack Flamp 的消息，但由于我的声望还没有 50，所以我不能。