c - 在 C 中 - for 循环中的 scanf() 导致 printf() 运行多次

Question

我正在完成一项作业，完成后，我有 1 个错误，并且修复了 1 个我不完全理解的错误。目前，只要用户按照要求进行操作，一切就可以正常工作。但我知道这种情况并不经常发生，所以我很想知道如何阻止这些问题。希望得到任何建议 - 我是 C 语言的初学者。

我在这里找到了许多不同的建议:C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf

我在 scanf() 语句中添加了一个空格，这解决了一些错误 - 我知道\n 被添加到输入的字符串/字符的末尾，我只是不确定如何检查它/处理它，我尝试使用 getchar() 代替 scanf() 但仍然遇到双重打印/循环问题。

错误问题

当用户运行游戏循环时，如果输入超过 1 个字符(例如:“oo”，当使用 scanf() 提示输入“y”或“n”时)，我的 printf 语句会运行 1x每个输入的字符，并相互连接:

示例是:

Welcome to Two doors.

Would you like to play? (y/n):Welcome to Two doors.

Would you like to play? (y/n):

如果用户输入“y”来玩游戏，但随后在第二部分中输入 1,2 或 3 以外的字符，也会出现此问题。

我如何限制他们的回复长度？或者是在输入 if 语句之前监视游戏长度和选择变量的最佳方法？也许检查它们是否长于 1 个字符，如果是，则只取第一个字符？

第二个问题 - 修复我不明白的错误在 scanf() 函数中，我遇到了与上面描述的非常相似的问题，但当用户输入任何字符时就会发生这种情况。我找到的解决方案是在字符前添加一个空格 ->

scanf(" %c", &play);

对比

scanf("%c", &play);

这个问题只有在使用循环时才会出现吗？因为在循环代码之前我从未发现这些错误。

使用 'while (getchar() != '\n');' 更新了代码Sourav Ghosh 的建议

#include <stdio.h>

int main(void) {

    char play;
    int choice;
    char answer[] = "No matter which one you choose the guards both tell you which door leads to death, and therefore you can pick the other door.\n";
    int gameLoop = 1;
    int timesPlayed = 0;

    while (gameLoop == 1){

        if (timesPlayed == 0) {
            printf("Welcome to Two doors.\n");
            printf("Would you like to play? (y/n):");
        } else {
            printf("Would you like to play again? (y/n):");
        }
        scanf(" %c", &play);
        while (getchar() != '\n');

        if (play == 'y') {
            // == instead of =
            printf("\nYou are a prisoner in a room with 2 doors and 2 guards.\n");
            printf("One of the doors will guide you to freedom and behind the other is a hangman --you don't know which is which.\n");
            printf("One of the guards always tells the truth and the other always lies. You don't know which one is the truth-teller or the liar either.\n");
            printf("You have to choose and open one of these doors, but you can only ask a single question to one of the guards.\n");
            printf("What do you ask so you can pick the door to freedom?\n\n");
            printf("\t1.Ask the truth-guard to point to the door of doom.\n");
            printf("\t2.Ask the liar-guard to point to the door of doom.\n");
            printf("\t3.Doesn't matter which one you pick.\n");
            scanf(" %d", &choice);
            while (getchar() != '\n');

            switch (choice) {

                case 1:
                printf("%s", answer);
                timesPlayed++;
                break;

                case 2:
                printf("%s", answer);
                timesPlayed++;
                break;

                case 3:
                printf("%s", answer);
                timesPlayed++;
                break;

                default:
                printf("The Troll Smasher comes out from the shadows and squeezes the stupid out of you until you pop. GAME OVER!\n");
                break;
            }
        } else if(play == 'n') {
            printf("Sorry to hear that, we at Two Doors hope you have a super duper day!\n");
            gameLoop = 0;
            break;
        } else {
            printf("That is not a valid input, please try again by entering either 'y' to start the game or 'n' to quit the game.\n");
        }
    }
    return 0;

}

Answer 1

%c 格式说明符的问题是，它只会从输入缓冲区读取一个字节，并且如果输入缓冲区存储了更多字节并且调用下次遇到时，它不会要求用户输入，它只会从可用输入流中读取下一个字节。

那么，回答一下

How can I limit the length of their response?

嗯，没有直接的方法可以阻止用户只输入 X 个字符/数字，而是刷掉多余的字符/数字(如果有)并下一次调用时，从空缓冲区开始是一种简单的方法。

因此，解决这个问题的快速方法是清除剩余输入的标准输入。你可以做类似的事情

  int retval = scanf(" %c", &play);

  //some code

  while (getchar() != '\n');   //eat up the input buffer

  //next call to scanf(), input buffer is empty now....

阻止 scanf() 读取已存在的不需要的输入，并强制它询问用户输入。

另外，不要忘记检查 scanf() 的返回值以确保调用成功。