Question

我正在尝试使“按q退出”功能正常工作，但是遇到了一些麻烦。 “无效字符”功能也无法正常工作。我将如何解决这个问题？我的教授建议在代码的开头放置“按q退出”功能，但没有给我任何进一步的说明。

public static void main(String[] args) {

    char userChar = 0;

    Scanner sc = new Scanner(System.in);
    Random rnd = new Random();

    // Intro/directions/prompting for user input
    System.out.println("Welcome to Rock, Paper, Scissors by Rancid!");
    System.out.println("Choose R for Rock, P for Paper, S for Scissors, or Q to Quit, then press Enter: ");

    // If player chooses to quit
    if (userChar == 'q' || userChar == 'Q') {
        System.out.println("Player chose to quit. Goodbye!");
    }

    // Start of loop
    while (userChar != 'q' && userChar != 'Q') {

        // Prompting computer to generate a random number
        int randomNumber = rnd.nextInt(3) + 1;

        // If computer generates 1 (Rock)
        if (randomNumber == 1) {
            if (userChar == 'r' || userChar == 'R') {
                System.out.println("Rock vs. Rock! It's a tie!");
            } else if (userChar == 'p' || userChar == 'P') {
                System.out.println("Paper covers Rock, you win!");
            } else if (userChar == 's' || userChar == 'S') {
                System.out.println("Rock breaks Scissors, you lose!");
            }
        }

        // If computer generates 2 (Paper)
        else if (randomNumber == 2) {
            if (userChar == 'r' || userChar == 'R') {
                System.out.println("Paper covers Rock, you lose!");
            } else if (userChar == 'p' || userChar == 'P') {
                System.out.println("Paper vs. Paper! It's a tie!");
            } else if (userChar == 's' || userChar == 'S') {
                System.out.println("Scissors cuts Paper, you win!");
            }
        }

        // If computer generates 3 (Scissors)
        else if (randomNumber == 3) {
            if (userChar == 'r' || userChar == 'R') {
                System.out.println("Rock breaks Scissors, you win!");
            } else if (userChar == 'p' || userChar == 'P') {
                System.out.println("Scissors cuts Paper, you lose!");
            } else if (userChar == 's' || userChar == 'S') {
                System.out.println("Scissors vs. Scissors! It's a tie!");
            }
        }

        // If player types an invalid character
        else {
            System.out.println("Invalid input! Please enter a valid character.");
        }

        userChar = sc.next().charAt(0);
        }
    }
}

Answer 1

好吧，首先，您要退出的条件超出了while循环的范围，这意味着它永远不会在给定的上下文中执行。

第二，除非userChar不同时是q和Q，否则while循环中的条件将不成立。

我建议您稍微重构一下while循环，例如：

while (true) {

    //Your logic here

    if (userChar == 'q' || userChar == 'Q') {
        System.out.println("Player chose to quit. Goodbye!");
        break;
    }


    //Your logic here
}

Answer 2

为了真正处理无效输入，您需要在循环中首先进行此操作。此外，您无法在else语句的if-else块中执行此操作，因为在那些if-else中，您检查的是randomNumber，而不是输入，因此它不会感觉，它将永远无法工作。另外，您检查输入内容是从sc.next().charAt(0)中获取的内容，但是例如，如果用户插入了"R_invalid_something_without_space"之类的字符串，您只会查看第一个字母R并将其当作有效输入。我修改了您的代码，以便您在每次循环迭代中读取一行，如果该行不是q / r / p / s，它将被视为无效输入，如果它的q，则该程序将结束。

 public static void main(String[] args) {

    char userChar = 0;

    Scanner sc = new Scanner(System.in);
    Random rnd = new Random();

    // Intro/directions/prompting for user input
    System.out.println("Welcome to Rock, Paper, Scissors by Rancid!");
    System.out.println("Choose R for Rock, P for Paper, S for Scissors, or Q to Quit, then press Enter: ");

    // Start of loop
    while (true) {
        // what I changed
        // **********************************************************
        String line = sc.nextLine().toLowerCase();
        if(line.equals("q")){
            System.out.println("Player chose to quit. Goodbye!");
            break;
        }
        if(!line.equals("r") && !line.equals("s") && !line.equals("p")) {
            System.out.println("Invalid input! Please enter a valid character.");
            continue;
        }
        userChar = line.charAt(0);
        // **********************************************************

        // Prompting computer to generate a random number
        int randomNumber = rnd.nextInt(3) + 1;

        // If computer generates 1 (Rock)
        if (randomNumber == 1) {
            if (userChar == 'r' || userChar == 'R') {
                System.out.println("Rock vs. Rock! It's a tie!");
            } else if (userChar == 'p' || userChar == 'P') {
                System.out.println("Paper covers Rock, you win!");
            } else if (userChar == 's' || userChar == 'S') {
                System.out.println("Rock breaks Scissors, you lose!");
            }
        }

        // If computer generates 2 (Paper)
        else if (randomNumber == 2) {
            if (userChar == 'r' || userChar == 'R') {
                System.out.println("Paper covers Rock, you lose!");
            } else if (userChar == 'p' || userChar == 'P') {
                System.out.println("Paper vs. Paper! It's a tie!");
            } else if (userChar == 's' || userChar == 'S') {
                System.out.println("Scissors cuts Paper, you win!");
            }
        }

        // If computer generates 3 (Scissors)
        else if (randomNumber == 3) {
            if (userChar == 'r' || userChar == 'R') {
                System.out.println("Rock breaks Scissors, you win!");
            } else if (userChar == 'p' || userChar == 'P') {
                System.out.println("Scissors cuts Paper, you lose!");
            } else if (userChar == 's' || userChar == 'S') {
                System.out.println("Scissors vs. Scissors! It's a tie!");
            }
        }
    }
}

剪刀石头布程序的某些方面无法正常运行

2 个答案: