Question

因此，对于这项任务，我必须再次包含一个游戏功能。意思是一旦人正确猜到该程序应该让用户选择是否再次玩。另外，我试图包含一个函数，如果用户在5次或更少的猜测中正确猜测，那么程序应该打印“Good Job！”如果它需要超过5个猜测，它应该显示“你可以做得更好！”。请帮帮我！我是编程的初学者，我一直试图解决这个问题。

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

int main( void )
{

int i, n = 0, r;
int answer;
srand( time( NULL ) );
r = rand() %100 +1;
char userName[15];


printf("Welcome to GUESS MY NUMBER!\n\nPlease type your name here: ");
scanf("%s", &userName);
printf("\n\nI am thinking of a number between 1 and 100.\n\nCan you guess what it is?  ");
while(scanf("%d", &i))
{      
if (n >= 9 && i != r)
{         
printf("\n\nSorry, the number was %d.\n", r);
printf("You should have gotten it by now.\n");        
printf("Better luck next time.\n\n");         
system ("PAUSE");
break;    
}      
if (i > r) 
{         
n++;
printf("Your guess is high. You only get 10 guesses. Try again: ");   
}     
else if (i < r) 
{         
n++;          
printf("Your guess is low. You only get 10 guesses. Try again: ");    
}     
else if (i == r) 
{             
printf("\n\nCongratulations %s!\nYou guessed the number within %d guesses!\nWould you  like to play again? y/n?\n",userName, n+1,answer);
scanf("%d", &answer);

system ("PAUSE");
break;         
}    
}
return 0;
}

Answer 1

一件简单的事情是创建一个bool变量（最初设置为true），可以在while语句中检查并在用户被赋予继续或不继续的选项后更新。然后将你的休息时间改为继续，你应该保持良好状态。

Answer 2

将整个事物包裹在另一个循环中，并在此外循环结束时，询问用户是否要再次播放。 while（）或do-while（）循环。如果用户说是，则继续循环，否则退出循环。

-Initialize the game
-Load any resources needed (in this case, none)

Begin looping continually
    - Handle input
    - Think
    - Show results
End looping if exited

-Free any resources (in this case, none)
-Exit

如何在我的C ++猜谜游戏中添加“再次播放”功能？循环麻烦

2 个答案: