用throw语句实现try-catch块

时间:2018-08-14 03:13:35

标签: java exception-handling try-catch throw

在我必须做的猜谜游戏中,我需要包含一个带有两个catch子句的try-catch块(一个针对我的两个自定义异常:BadGuessExceptionTooManyGuessesException的多重catch块,以及一个NumberFormatException阻止)。

我尝试在程序中创建条件以引发自定义异常,因为我不知道它们如何正常工作背后的逻辑。我遇到了编译错误,并希望帮助重新设计我的程序,以便它正确地实现try-catch-catch块。

我的自定义异常类:

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!");
  }

}

我的程序出现编译错误:

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

public class GuessingGame throws NumberFormatException
{
  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;

    //Initialize incrementor for guessing turns
    int turn = 1;

    //Create variable for user input (guess)
    int guess;

    try
    {
      while(guess != n)
      {
        //Exception handling for more than five turns
        if(turn > 5)
          throw new TooManyGuessesException();

        //Prompt user to enter their guess
        System.out.println("Guess a number between 1 and 10 inclusive.");
        System.out.println("Hint: the answer is " + n);
        //Receive user input (their guess)
        guess = keyboard.nextInt();

        //Increment turn variable
        turn++;

        if(guess < 1 || guess > 10)
          throw new BadGuessException();
        else if(guess == n)
          System.out.println("YOU WIN!\nIt took you " + turn + " attempts.");
      }
    }
    catch(BadGuessException e | TooManyGuessesException e)
    {
      e.getMessage();
    }
    catch(NumberFormatException e)
    {
      System.out.println("Sorry, you entered an invalid number format.");
    }

  }
}

1 个答案:

答案 0 :(得分:1)

进行GuessingGame类中的更改在多个try块中,在BadGuessException之后删除 e 。并用0初始化猜测,并从类声明中删除NumberformatException;

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;

    //Initialize incrementor for guessing turns
    int turn = 1;

    //Create variable for user input (guess)
    int guess = 0 ;

    try
    {
      while(guess != n)
      {
        //Exception handling for more than five turns
        if(turn > 5)
          throw new TooManyGuessesException();

        //Prompt user to enter their guess
        System.out.println("Guess a number between 1 and 10 inclusive.");
        System.out.println("Hint: the answer is " + n);
        //Receive user input (their guess)
        guess = keyboard.nextInt();

        //Increment turn variable
        turn++;

        if(guess < 1 || guess > 10)
          throw new BadGuessException();
        else if(guess == n)
          System.out.println("YOU WIN!\nIt took you " + turn + " attempts.");
      }
    }
    catch(BadGuessException | TooManyGuessesException e)
    {
      e.getMessage();
    }
    catch(NumberFormatException e)
    {
      System.out.println("Sorry, you entered an invalid number format.");
    }

  }
}