scanf无法处理无效的输入 [英] scanf not working on invalid input

问题描述

在第一个 scanf（）中输入字符时，第二个字符不运行。 getchar（）不能用于重试输入。跳过输入您是否想再次播放？ （是/否）？看来 your_choice 应该是字符，然后再检查，但是字符实际上是由 ch 。是什么导致它像这样工作以及如何解决问题。我试过重新初始化变量，但不起作用。

On a character input in the first scanf(), the second one doesn't run. getchar() isn't working either for Try Again input. It skips to take input for Would you like to play again? (Y/N)? It seems that your_choice is supposed to take the character and check it afterward but the character is actually being taken by ch. What is causing it to work like this and how to resolve the issue. I've tried re-initializing the variables but doesn't work.

#include <stdio.h>

void choice(int);

int main() {
    char ch;
    int random, your_choice;

    do {
        srand(time(NULL));
        system("cls");
        printf("** 0 is for Rock **\n");
        printf("** 1 is for Scissors **\n");
        printf("** 2 is for Lizard **\n");
        printf("** 3 is for Paper **\n");
        printf("** 4 is for Spock **\n");

        printf("\nEnter your choice here:");
        scanf("%d", &your_choice);

        random = rand() % 5; //random number between 0 & 4
        if ((your_choice >= 0) && (your_choice <= 4)) {
            //choice printer omitted for this post

            if ((random == ((your_choice + 1) % 5)) || (random == ((your_choice + 2) % 5)))
                printf("\n\n... and you win!!!\n");
            else if ((random == ((your_choice + 3) % 5)) || (random == ((your_choice + 4) % 5)))
                printf("\n\n... and you lose!!!\n");
            else if (random == your_choice)
                printf("\n\nUnfortunately, it's a tie!\n");
        } else
            printf("\nWell, this is wrong! Try again with a number from 0 to 4!!\n");

        printf("\nWould you like to play again? (Y/N)?: ");
        scanf(" %c", &ch);

    } while (ch == 'y' || ch == 'Y');

    return 0;
}

推荐答案

如果用户输入的字符无法转换为数字， scanf（％d，& your_choice）; 返回0，而 your_choice 为保持不变，因此未初始化。行为是不确定的。

If the user enters characters that cannot be converted to a number, scanf("%d", &your_choice); returns 0 and your_choice is left unmodified, so it is uninitialized. The behavior is undefined.

您应该对此进行测试并以这种方式跳过有问题的输入：

You should test for this and skip the offending input this way:

    if (scanf("%d", &your_choice) != 1) {
        int c;
        /* read and ignore the rest of the line */
        while ((c = getchar()) != EOF && c != '\n')
            continue;
        if (c == EOF) {
            /* premature end of file */
            return 1;
        }
        your_choice = -1;
    }

说明：

• scanf（）返回成功的转换次数。如果用户键入数字，则将其转换并存储到 your_choice 中，如果用户输入 scanf（），则返回1输入不是数字的内容，例如 AA ， scanf（）将有害的输入保留在标准输入缓冲区中并返回0，最后如果到达文件末尾（用户在Windows中输入^ Z输入或在UNIX中输入^ D）， scanf（）返回 EOF 。

• scanf() returns the number of successful conversions. If the user types a number, it is converted and stored into your_choice and scanf() returns 1, if the user enters something that is not a number, such as AA, scanf() leaves the offending input in the standard input buffer and returns 0, finally if the end of file is reached (the user types ^Z enter in windows or ^D in unix), scanf() returns EOF.

如果输入未转换为数字，则输入的正文if 语句：使用 getchar（）一次消耗一个字节的输入，直到读取文件末尾或换行符为止。

if the input was not converted to a number, we enter the body of the if statement: input is consumed one byte at a time with getchar(), until either the end of file or a linefeed is read.

如果 getchar（）返回 EOF ，我们已经读取了整个输入流，无需提示用户进行更多输入，您可能需要在返回错误代码之前输出错误消息。

if getchar() returned EOF, we have read the entire input stream, no need to prompt the user for more input, you might want to output an error message before returning an error code.

否则，将 your_choice 设置为 -1 ，该值无效，因此读取

otherwise, set your_choice to -1, an invalid value so the read of the code complains and prompts for further input.

必须读取并丢弃有问题的输入：如果您不这样做，这样做，下一个输入语句 scanf（％c，& ch）; 将读取有问题的输入的第一个字符，而不是等待用户输入以响应您想再玩一次吗？ （是/否）？：提示。这是您观察到的行为的解释。

Reading and discarding the offending input is necessary: if you do not do that, the next input statement scanf(" %c", &ch); would read the first character of the offending input instead of waiting for user input in response to the Would you like to play again? (Y/N)?: prompt. This is the explanation for the behavior you observe.

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