石头剪刀纸程序无法正常运行的方面

Question

我试图让“按q退出”功能正常工作，但我遇到了一些麻烦。“无效字符”功能也无法正常工作。我该如何解决这个问题？我的教授建议在我的代码开头放置“按q退出”函数，但没有给我任何进一步的指导。

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!");

            }

        }

Answer 1

为了真正处理无效输入，您需要在循环中首先完成此操作。此外，你不能在你的语句块中这样做，因为在那些你检查的语句中，而不是输入，所以它没有意义，它永远不会起作用。此外，您将输入检查为从中获取的内容，但是例如，如果用户插入一个字符串，例如您只查看第一个字母，然后将其作为有效输入。我修改了你的代码，以便你在每次循环迭代中读取一行，如果这一行不是，它将被视为无效输入，如果它的程序将完成。elseif-elseif-elserandomNumbersc.next().charAt(0)"R_invalid_something_without_space"Rq / r / p / sq 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!");            }        }    }}

Answer 2

好吧，首先，您退出的条件超出了 while 循环的范围，这意味着它永远不会在此给定上下文中执行。其次，while 循环中的条件将不为 true，除非 userChar 不是同时是 q 和 Q，这是没有意义的。我的建议是稍微重构一下你的 while 循环，如下所示：while (true) {    //Your logic here    if (userChar == 'q' || userChar == 'Q') {        System.out.println("Player chose to quit. Goodbye!");        break;    }    //Your logic here}