随机数字游戏C

Question

虽然我的第一次游戏运行良好，第二次它只给你两条生命......我试图改变生命的数量，但仍然无法弄清楚我做错了什么。

// C_program_random_number_game

#include<stdio.h> 
#include<time.h> 
#include <stdlib.h> 

int main() 
{ 

    srand(time(NULL)); 
    int num1,x = 0; 
    char game, cont, replay; 

    printf("Would you like to play a game? : "); 
    scanf("%c",&game); 

    if (game == 'y' || game == 'Y') 
    { 
    printf("\nThe rules are simple. You have have 5 tries to guess the computers number. \n \n If you succeed you win the game, if you dont you lose the game. Good luck!"); 
    do 
    { 
     int r = rand()%5 +1; 
     printf("\n\nEnter a number between 1 and 5 : "); 
     scanf("\n%d",&num1); 
     x++; 
     if(num1 > 0 && num1 < 5) 
     { 

      do 
      { 
      if(num1 < r) 
       { 
        printf("\nClose! try a little higher... : "); 
        x++; 
       } 
       else if (num1 > r) 
       { 
        printf("\nClose! try a little lower...: "); 
        x++; 
       } 
       scanf("%d",&num1); 

      }while(num1!=r && x <3); 

      if(num1 == r) 
      { 
       printf("\nWinner! >> you entered %d and the computer generated %d! \n",num1, r); 
      } 
      else if(num1 != r) 
      { 
       printf("\nBetter luck next time!"); 
      } 
      printf("\n\nWould you like to play again? (y or n) : "); 
      scanf("\n%c",&replay); 
     } 
      else 
     { 
      printf("Sorry! Try again : "); 
      scanf("%d",&num1); 
     } 

    }while(replay == 'y'|| replay == 'Y'); 

} 
else if (game == 'n' || game == 'N') 
{ 
    printf("Okay, maybe next time! "); 
} 
else 
{ 
    printf("Sorry, invalid data! "); 
} 
return 0; 

}

Answer 1

有你的代码有各种问题（其中大多数是在编程方面很小）。大部分错误都是你想通过这个问题和你的printf（）完成的。

因为这个代码将在1-25之间随机选择，接受任何有效int的输入，看看你是否匹配它，只给你5次尝试。 （我没有添加错误检查来强制输入介于1-25之间，这应该可以添加。）

我评论了我的代码，下面列出了我所做的所有更改，并通过printf（）秒。

注意：由于我已经提出了自己的意见，请参阅上面的意见以了解我所做的更改。我也格式化了它，使它更容易阅读。

注2：我使用在线编译器很快完成了这个任务。如果你发现有什么问题或者不能按你的意愿工作，请在下面评论，我会解决它。

// C_program_random_number_game 

#include<stdio.h> 
#include<time.h> 
#include <stdlib.h> 

int main() 
{ 

    srand(time(NULL)); 
    int num1,x = 0; 
    char game, cont, replay; 

    printf("Would you like to play a game? : "); 
    scanf("%c",&game); 

    if (game == 'y' || game == 'Y') 
    { 
     printf("\nThe rules are simple. You have have 5 tries to guess the     computers number. \n \n If you succeed you win the game, if you dont you lose the game. Good luck!"); 

     do 
     { 
      int r = rand()%25 +1; 

      printf("\n\nEnter a number between 1 and 25 : "); 
      scanf("%d",&num1); 

      do 
      { 
       printf("r = %d\n", r); 

       if(num1 < r) 
       { 
        printf("\nClose! try a little higher... : "); 
        x++; //Increment x if wrong guess 
       } 
       else if (num1 > r) 
       { 
        printf("\nClose! try a little lower...: "); 
        x++; //Increment x if wrong guess 
       } 

       scanf("%d",&num1); 
      }while(num1!=r && x < 5); //If x is 5 or more, they ran out of guesses (also, you want an && not an ||) 

      if(num1 == r) //Only print "winner" if they won! 
      { 
       printf("\nWinner! >> you entered %d and the computer generated %d! \n",num1, r); 
      } 

      printf("\nWould you like to play again? (y or n) : "); 
      scanf("\n%c",&replay); 
     }while(replay == 'y'|| replay == 'Y'); 
    } 

    printf("Thanks for playing! "); 

    if (game == 'n' || game == 'N') 
    { 
     printf("Okay, maybe next time! "); 
    } 
    return 0; 
} 

Answer 2

有两个问题的组合。首先，当数字匹配时，你不会摆脱“for”循环。因此，只有每第三次猜测才会检查比赛。

第二个问题是在这样的逻辑检查：

}while(num1!=r || x <= 3); 

我们看到，这个变成“真”，如果在for循环被打破了早。