Ç - 试图读取从标准输入一个字符（和失败）W / scanf函数/的getchar [英] C - trying to read a single char from stdin (and failing) w/ scanf / getchar

问题描述

作为家庭作业的一部分，我想从stdin读取一个单个字符，并采取行动根据它：

as a part of a homework assignment, I'm trying to read a single char from stdin, and act according to it:

char choice;

while (1) {
    printf("please enter [y]es, [n]o or [m]aybe: ");
    scanf("%c", choice);
    fflush(stdin);
    // validate input
    if (choice == 'y' || choice == 'n' || choice == 'm') {
        break;
    } else {
      printf("Please enter only \'y\', \'n\' or \'m\'!\n");
    }
}
// do something with the input
if (choice == 'y') {
    printf("you selected yes!\n");
}

由于某种原因，scanf的同时捕捉焦炭，换行之后，因此，它继续做与输入的东西，然后同时打印请输入只有'Y'，'n'或'M'！线。如果我在标准输入输入几个字符，将打印该行对所有的人，同时也是第一个字符的正确执行。因此，例如：

for some reason, scanf captures both the char and the line-feed after, thus it proceeds to do something with the input and then also prints the "Please enter only 'y', 'n' or 'm'!" line. If I enter several characters on the stdin, it will print that line for all of them, while also performing correctly for the first char. So, for example:

$ ./run
please enter [y]es, [n]o or [m]aybe: y<return>
you selected yes!
Please enter only 'y', 'n' or 'm'!
$ ./run
please enter [y]es, [n]o or [m]aybe: yes<return>
you selected yes!
Please enter only 'y', 'n' or 'm'!
Please enter only 'y', 'n' or 'm'!
Please enter only 'y', 'n' or 'm'!
$

如果我使用的getchar同样的事情发生。我在想什么？谢谢。

Same thing happens if I use getchar. What am I missing? thanks.

推荐答案

您需要scanf函数（和之间的空间％C 为它正常工作：

You need a space between scanf(" and the %c for it to work correctly:

scanf(" %c", &choice);

而且你还需要使用＆安培;！选择，而不是选择

And you also need to use &choice, not choice!

编辑：当你在这，你可能想看看做的，而（）为循环（除非教授明确表示使用破）！ - 做的，而验证用户输入时的伟大工程

While you're at it, you might want to look into do while() for that loop (unless the professor specifically said to use a break) - do while works great when validating user input!

这篇关于Ç - 试图读取从标准输入一个字符（和失败）W / scanf函数/的getchar的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！