首先与fgets（）调用输入时被跳过？ [英] First fgets() call being skipped during input?

问题描述

我在写一个程序，可以采取计算机细节在手动模式下，用户。不过，我碰到了一点小问题。

I'm writing a program that can take computer specifics from the user in manual mode. However, I've run into a bit of a problem.

在此code：

    char choice = getc(stdin);

    if (choice == 'y' || choice == 'Y')
    {
        config_file = fopen(config_file_loc, "w");

        printf("%s", "Please enter the name of your distribution/OS: ");
        fgets(distro_str, MAX_STRLEN, stdin);
        fputs(distro_str, config_file);
        fputs("\n", config_file);

        printf("%s", "Please enter your architecture: ");
        fgets(arch_str, MAX_STRLEN, stdin);
        fputs(arch_str, config_file);
        fputs("\n", config_file);

        fclose(config_file);
    }

在运行时，输入正确的跳跃，从请输入您的发行/ OS的名称：来请输入您的架构：留下distro_str空白

During runtime, the input jumps right from "Please enter the name of your distribution/OS:" to "Please enter your architecture:", leaving distro_str blank.

我试过冲洗stdin和stdout，但这些都没有奏效。

I've tried flushing stdin and stdout, but those haven't worked.

感谢您的帮助。

推荐答案

当你调用 GETC（标准输入）来获得选字，它读取一个字符从输入缓冲器。如果用户输入Y后跟一个换行符，则换行符将保留在输入缓冲区，以及随后的与fgets（）通话将读取 - 一个空行 - 并立即返回。

When you call getc(stdin) to get the "choice" character, it reads exactly one character from the input buffer. If the user entered "y" followed by a newline, then the newline will remain in the input buffer, and the subsequent fgets() call will read that - an empty line - and return immediately.

如果您希望所有输入要面向行的，那么你应该叫与fgets（）每次看完选择的价值时，包括

If you wish all input to be line-oriented, then you should call fgets() each time, including when reading the "choice" value.

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