简单的随机数游戏程序

Question

这是我的代码，这是一个简单的数字游戏，用户试图猜测随机数，但是我不知道为什么你永远不会赢。有两件事我无法解决 1) 用户永远猜不出正确的数字 2) 我想猜猜 3 次，尽管我似乎无法抓住我忽略的内容

// C_program_random_number_game

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




int main(){

srand(time(NULL));
int num1,x;
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 5 : ");
    scanf("%d",&num1);
    do{
    for(x=1; x<=3; x++){

    if(num1 < r){
        printf("\nClose! try a little higher... : ");
    }
    else if (num1 > r){
        printf("\nClose! try a little lower...: ");
    }
    scanf("%d",&num1);
    }
    }while(num1!=r || x <= 3);
    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 1

一个明确的问题是格式说明符不正确：

scanf("&d",&num1);

应该是：

scanf("%d",&num1);

此外，您的while 循环中的最后两个条件永远不会被评估，因为如果猜测等于随机数，它就不会进入循环。使用do-while 循环代替无限 循环，并根据用户输入将break 循环出来。请记住在循环中获取用户输入的猜测。

Answer 2

有各种缺陷，这里是你要找的代码：

while(num1 != r){
    if(num1 < r){
        printf("higher... : ");
    }
    else{
        printf("lower...: ");
    }
    scanf("%d",&num1);
}
printf("Winner! >> you entered %d and the computer generated %d! \n",num1, r);
printf("Would you like to play again? (y or n) : ");
scanf("%c",&replay);

正如我在上面的评论中指出的，之前在 while 循环中，永远不会有 num1==r 的时候，因此里面的 if 语句永远不会是真的。现在，循环 jsut 在到达数字后就停止了。