随机猜数游戏的逻辑错误

Question

在这个程序中，计算机生成一个随机数（在1-100之间），用户试图猜测它。一直运行到用户正确猜出数字为止。需要打印出在正确猜测数字之前所花费的尝试总数。程序运行正常，但存在逻辑错误。当我正确猜测数字时，什么都没有打印出来;程序只是停止。

import java.util.*;
public class Problem3 {


  public static void main(String[] args) { 

    Random rand = new Random();
    Scanner console = new Scanner(System.in);

    int num = rand.nextInt(100)+1;
    System.out.println(num); //prints out the random number so I know test works correctly
    int count = 0;
    System.out.println("Guess the random number. The number is between 1 and 100. Enter it below");
    console = new Scanner(System.in);
    int guess = console.nextInt();
    while(guess != num){  //while guess is not = to number  
      if(guess < num){             //if less than num
        System.out.println("Your guess is too low"); 
        System.out.println("Guess the random number. The number is between 1 and 100. Enter it below");
        guess = console.nextInt();
        count++;
      }else if( guess > num){      //if greater than num   
          System.out.println("Your guess is too high");
          System.out.println("Guess the random number. The number is between 1 and 100. Enter it below");
          guess = console.nextInt();
          count++;
      }else{
        count++;
        System.out.println("You guessed correctly after " + count + " tries!");
    }
    }
  }

}

Answer 1

当猜到正确的数字时，while循环条件变为false。这会使其他人无法访问。

编辑

下面是我如何解决你的问题。如果你发现自己在重复代码，这通常意味着有改进的机会。不仅代码更少，而且更容易阅读和理解正在发生的事情。我并不是试图窃取答案（@AmirBawab首先发现了问题），但我坚信不仅要编写有效的代码，还要编写可以由他人维护和阅读的代码。希望当你继续学习时，你会记住这一点。编码愉快！

public static void main(String[] args) {

    Random rand = new Random();
    Scanner console = new Scanner(System.in);

    int num = rand.nextInt(100) + 1;

    int count = 0;

    while (true) {
        System.out.println("Guess the random number. The number is between 1 and 100. Enter it below");
        int guess = console.nextInt();
        count++;
        if (guess < num) {
            System.out.println("Your guess is too low");
        } else if (guess > num) {
            System.out.println("Your guess is too high");
        } else {
            System.out.println("You guessed correctly after " + count + " tries!");
            break;
        }
    }

    console.close();

}

Answer 2

实际上，您永远不会进入< code>else阶段，因为当数字被猜出时，< code>while代码不会执行，因此< code>else代码也永远不会执行。在此之后，数字被猜测，因此条件为< code>false，退出< code>while循环，并< code>System.out.print祝贺消息

将它放在while循环的“末尾”:

 System.out.println("You guessed correctly after " + count + " tries!");

下面是您的代码应该是什么样子的：

while(guess != num){  //while guess is not = to number  
      if(guess < num){             //if less than num
        System.out.println("Your guess is too low"); 
        System.out.println("Guess the random number. The number is between 1 and 100. Enter it below");
        guess = console.nextInt();
        count++;
      }else{      //if greater than num   
          System.out.println("Your guess is too high");
          System.out.println("Guess the random number. The number is between 1 and 100. Enter it below");
          guess = console.nextInt();
          count++;
      }
    }

  System.out.println("You guessed correctly after " + ++count + " tries!");