C - 添加额外步骤时执行

Question

当我运行下面的代码时，它会正确执行，但会在我再次输入之前添加一个额外的迭代。不知道为什么。我将使用它来为孩子分叉和管道父进程。该步骤将在用户输入进程数之后进行。

代码：

#include <stdio.h> 

int main(void) {
    int num_processes;
    int running = 1;
    do {
        printf ("How many processes do you want? (1, 2 or 4) \n");
        num_processes = getchar();
        num_processes -= '0';

        if(num_processes == 1 || num_processes == 2 || num_processes == 4){
            int temp = num_processes - 1;
            printf("\n1 Parent and %d child processes\n", temp);
            printf("----------------------------------\n");
            running = 0;
        } else {
            printf("Invalid Input, please try again.\n");
        }
    } while(running == 1);

    // Do important stuff

    return(0);
}

输出：

How many processes do you want? (1, 2 or 4)
3
Invalid Input, please try again.
How many processes do you want? (1, 2 or 4)
Invalid Input, please try again.
How many processes do you want? (1, 2 or 4)
2

1 Parent and 1 child processes
----------------------------------

Answer 1

想想当你给出第一个输入时会发生什么。您输入 3，然后按 Enter。

现在，在标准输入缓冲区中，有两个项目等待使用，3 和 \n（即 Enter）。

你第一次进入循环体，消耗了 3。现在下一个排队等待\n...

当您再次（第二次）进入循环体时，标准输入缓冲区要读取的下一个字符是\n，这就是第二个getchar() 愉快地返回给您的字符。

这是一个快速修复：

do {
    printf ("How many processes do you want? (1, 2 or 4) \n");
    // Eat trailing newline character, if any
    do {
      num_processes = getchar();      // read a character
    } while(num_processes == '\n');   // if it was the newline character, repeat
                                      // until `num_processes` is not the newline character

    // Continue with your logic..
    num_processes -= '0';

输出：

How many processes do you want? (1, 2 or 4) 
3
Invalid Input, please try again.
How many processes do you want? (1, 2 or 4) 
2

1 Parent and 1 child processes
----------------------------------

PS：正如@Barmar 评论的那样，如果您想阅读行，请使用fgets()。在这里阅读更多：How to use fgets() to control the execution of while loop through user input in c?