Question

嗨，我目前在猜测游戏代码时遇到异常。我想为字符串输入（而不是int）添加异常，并且还要输入超出限制（50）的数字的例外。感谢。

public static void main (String[] args)

{
    Random ran = new Random();
    int numberToGuess = ran.nextInt(50)+1;
    int numberOfTries = 0;
    Scanner input = new Scanner (System.in);
    int guess;
    boolean win = false;

    while (win == false)
    {
        System.out.println("Guess a number from 1 to 50!");
        guess = input.nextInt();
        numberOfTries++;

        if (guess == numberToGuess)
        {
            win = true;
        }
        else if (guess < numberToGuess)
        {
            System.out.println("Too low. Try again.");
        }
        else if (guess > numberToGuess)
        {
            System.out.println("Too high. Try again.");
        }
    }

    System.out.println("You got it in " + numberOfTries + " attempt(s)!");

}

Answer 1

这是一个潜在的解决方案：

import org.apache.commons.lang3.StringUtils;

import java.util.Random;
import java.util.Scanner;

public class StackOverflow45419907 {
  private final Scanner input;
  final int numberToGuess;

  public StackOverflow45419907() {
    this.input = new Scanner(System.in);
    numberToGuess = new Random().nextInt(50) + 1;
  }

  public static void main(String[] args) {
    new StackOverflow45419907().playGame();
  }

  private void playGame() {
    int numberOfTries = 0;
    int guess = -1;

    while (guess != numberToGuess) {
      guess = collectGuess();
      numberOfTries++;

      printClue(guess);
    }
    System.out.println("You got it in " + numberOfTries + " attempt(s)!");
  }

  private void printClue(int guess) {
    if (guess < numberToGuess) {
      System.out.println("Too low. Try again.");
    } else if (guess > numberToGuess) {
      System.out.println("Too high. Try again.");
    }
  }

  private int collectGuess() {
    System.out.println("Guess a number from 1 to 50!");
    final String potentialGuess = input.nextLine();
    return validateAndParse(potentialGuess);
  }

  private int validateAndParse(String potentialGuess) {
    if (!StringUtils.isNumeric(potentialGuess)) {
      throw new IllegalArgumentException("not numeric: " + potentialGuess);
    }
    final int asInt = Integer.parseInt(potentialGuess);
    if (asInt > 50 || asInt < 1) {
      throw new IllegalArgumentException("value out of valid range: " + asInt);
    }
    return asInt;
  }
}

Answer 2

我真的认为你不需要抛出异常。您可以使用 Scanner#hasNextInt()验证输入是否为整数。然后将它分配给猜测变量，然后检查它是否大于50。

如果你真的要抛出异常。使用 throw new RuntimeException(message)如果输入不是整数或大于50。

修改我不会那样使用它，但我相信你只想了解异常。

System.out.println("Guess a number from 1 to 50!"); numberOfTries++; if (!input.hasNextInt()) throw new IllegalArgumentException("Input must be a number"); guess = input.nextInt(); if (guess < 1 || guess > 50) throw new IllegalArgumentException("Input must be a number from 1 to 50");

Java猜测游戏的异常放在哪里

2 个答案: