Question

我是编码新手。我创建了一个猜谜游戏，并且运行良好，但是，我想知道如何制作它，以便在用户尝试对数字进行3次猜测之后，他们会得到提示，我将其放在最后一行，但目前无法实现，而且我不知道如何在train = df.loc[perm[:train_end]] test = df.loc[perm[train_end:]] #include "C:\Users\CLOS\Desktop\New folder\Reg9s12.H" TEMP DC.B 1 RAMEnd EQU $1001 MAIN: LDS #RAMEnd+1 BSET INTCR,%1100000 CLR TEMP CLI BSET DDRB,%00010000 BCLR PORTB,%00010000 HERE BRA HERE IRQ_EDGE LDAA TEMP EORA #%00010000 STAA TEMP RTI ORG $FFF2 DC.W IRQ_EDGE ORG $FFFE DC.W MAIN循环中使语句可访问。我目前陷入困境。谢谢

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Answer 1

程序中存在一些流程问题，但这是一种修复程序的简单方法。

首先，当您从guess方法中返回run()的值时，实际上并没有使用它，因此可以将其删除。

此外，在这种情况下，您不想使用do/while循环，而只想使用while。您要一直重复提示，直到用户正确猜出为止。因此，添加boolean可以检查他们是否赢了：

boolean correct = false;-我们将其设置为false，因为他们还没有赢。

现在，不要在每次猜测后再次调用run()（每次都会重置tries计数，只需让while循环执行其工作并重复一次即可。因此，我们需要将提示输入移至while循环中。

以下是更改的完整代码清单：

import java.util.Scanner;

public class guessing_game {

    public static void main(String[] args) {
        Scanner kb = new Scanner(System.in);
        desc();
        run(kb);

        //int nun = 0;    
        //for (int i = 0; i < nun; nun ++)   
    }

    public static void desc() {
        System.out.println("This is a guessing game.");
        System.out.println();
        System.out.println("Let's see how many tries it takes you to guess the right number!");
        System.out.println();
        System.out.println();
        System.out.println();
    }    

    // Change the return type to void as you never use the value returned
    public static void run(Scanner kb) {    
        int num = 44;

        // Add a boolean to determine if the game is won
        boolean correct = false;

        int tries = 0;

        while (!correct) {

            System.out.println("Please enter a number between 1-100");
            int guess = kb.nextInt();

            if (guess < num) {
                System.out.println("Oooh. Your guess is too low. Try again.");
                System.out.println();
            } else if ((guess > 100) || (guess < 0)) {
                System.out.println("That isn't between 1-100 is it?");
                System.out.println();
            } else if (guess > num) {
                System.out.println("Aaah. Your guess is too high. Try again.");
                System.out.println();
            } else if (guess == num) {

                // Flag the guess as correct; this will exit the loop after this run
                correct = true;

                System.out.println("Bingo!!! Nice guess bud.");
                System.out.println("Tell a friend to play! Wanna try again? (y or n)");
                String choice = kb.next();
                if (choice.equalsIgnoreCase("y")) {
                    run(kb);

                } else if (choice.equalsIgnoreCase("n")) {
                    System.exit(0);
                }
            }
            tries++;
        }    
    }
}

Answer 2

1。您递归地调用run（）方法，每次调用此方法时，都会创建一个新的变量try并将其初始化为零。 2.您的递归调用在条件检查之前，并且由于同样的原因，逻辑可能永远也不会到达条件检查。

要以最小的更改使它起作用，可以使用以下代码。但这不是最好的方法，因为它不能解决上述缺点

import java.util.Scanner;

公共类的guessing_game {

static int tries = 0;

   public static void main (String[] args)
   {
      Scanner kb = new Scanner(System.in);
      desc();
      run(kb);    

       //int nun = 0;

       //for (int i = 0; i < nun; nun ++)


   }
   public static void desc()
   {
      System.out.println("This is a guessing game.");
       System.out.println();
       System.out.println("Let's see how many tries it takes you to guess the right number!");
       System.out.println();
       System.out.println();
       System.out.println();
   }


   public static int run(Scanner kb)
   {
       System.out.println("Please enter a number between 1-100");
       int guess = kb.nextInt();

       int num = 44;


       do{

           tries++;
           if(tries>=3)  break;

       if (guess < num)
         {
            System.out.println("Oooh. Your guess is too low. Try again.");
            System.out.println();
            run(kb);
         }
       else if ((guess > 100) || (guess < 0))
         {
            System.out.println("That isn't between 1-100 is it?");
            System.out.println();
            run(kb);
         }

       else if (guess > num)
         {
            System.out.println("Aaah. Your guess is too high. Try again.");
            System.out.println();
            run(kb);
         }
       else if(guess == num)
         {
            System.out.println("Bingo!!! Nice guess bud.");
            System.out.println("Tell a friend to play! Wanna try again? (y or n)");
            String choice = kb.next();
            if (choice.equalsIgnoreCase("y"))
               {
                  run(kb);

               }
            else if (choice.equalsIgnoreCase("n"))
               {
                  System.exit(0);
               }
         }

         }while( tries < 3);
            {
               System.out.print("Here's a hint the lucky number is 4");
            }

           return guess;
   }




}

经过一定次数的尝试后打印打印语句

2 个答案: