c - 我的 Do-While 循环似乎没有循环？

Question

请帮忙，当我要求用户输入 continue using scanf 和 printf 时，我似乎无法让 do-while 循环循环。可能是 wipe_buffer 吗？我似乎无法在这方面做对，我只需要完成循环，这应该主要是功能性的。编码新手和这个网站请不要太苛刻。请谢谢。

    #include <stdio.h>
    #include <stdlib.h>
void wipe_buffer(void);
int main(int argc, char* argv[])

{
    char play1;
    char play2;
    char cont;

    do{
        printf("Player one pick Rock, Paper, or Scissors\n");
        scanf(" %c", &play1);
        wipe_buffer();
        printf("Player two pick Rock, Paper, or Scissors\n");
        scanf(" %c", &play2);
        wipe_buffer();

        switch(play1)
        {
            case 'r':
                 if(play2 == 'p' || play2 == 'P')
            {
                printf("Paper Covers Rock\n");
                printf("Player two wins\n");
            }
                if(play2 == 's' || play2 == 'S')
                {
                    printf("Rock Breaks Scissors\n");
                    printf("Player one wins\n");
                }
                if(play2 == 'r' || play2 == 'R')
                {
                    printf("Draw, Nobody wins\n");
                }
            break;
            case 'R':
                if(play2 == 'p' || play2 == 'P')
            {
                printf("Paper Covers Rock\n");
                printf("Player two wins\n");
            }
                if(play2 == 's' || play2 == 'S')
                {
                    printf("Rock Breaks Scissors\n");
                    printf("Player one wins\n");
                }
                if(play2 == 'r' || play2 == 'R')
                {
                    printf("Draw, Nobody wins\n");
                }
            break;
            case 'P':
                 if(play2 == 'p' || play2 == 'P')
            {
                printf("Draw, Nobody wins\n");
            }
                if(play2 == 's' || play2 == 'S')
                {
                    printf("Scissors cuts Paper\n");
                    printf("Player two wins\n");
                }
                if(play2 == 'r' || play2 == 'R')
                {
                    printf("Paper covers rock\n");
                }
            break;
            case 'p':
                 if(play2 == 'p' || play2 == 'P')
            {
                printf("Draw, Nobody wins\n");
            }
                if(play2 == 's' || play2 == 'S')
                {
                    printf("Scissors cuts Paper\n");
                    printf("Player two wins\n");
                }
                if(play2 == 'r' || play2 == 'R')
                {
                    printf("Paper covers rock\n");
                }
            break;
            case 's':
                 if(play2 == 'p' || play2 == 'P')
            {
                printf("Scissors Cuts Paper\n");
                printf("Player one wins\n");
            }
                if(play2 == 's' || play2 == 'S')
                {
                    printf("Draw, Nobody wins\n");
                }
                if(play2 == 'r' || play2 == 'R')
                {
                    printf("Rock breaks Scissors\n");
                    printf("Player two wins\n");
                }
            break;
            case 'S':
                 if(play2 == 'p' || play2 == 'P')
            {
                printf("Scissors Cuts Paper\n");
                printf("Player one wins\n");
            }
                if(play2 == 's' || play2 == 'S')
                {
                    printf("Draw, Nobody wins\n");
                }
                if(play2 == 'r' || play2 == 'R')
                {
                    printf("Rock breaks Scissors\n");
                    printf("Player two wins\n");
                }
            break;
        }
    printf("Do you wish to continue?\n");
    scanf(" &c", &cont);
    wipe_buffer();
    }while(cont == 'y' || cont == 'Y');
}
void wipe_buffer(void)
{
    char t;
    scanf("%c", &t);
    while(t != '\n' )
    {
        scanf("%c", &t);
    }
    return;
}

Answer 1

在本节中:

    printf("Do you wish to continue?\n");
scanf(" &c", &cont);

"&c" 应该是 "%c"

请注意，当我使用 gcc 编译它时，我收到一条基本上解决了这个问题的警告，因此请务必阅读您的警告。