Question

在提出这个问题之前，我读过：fgetc(stdin) in a loop is producing strange behaviour。我正在编写一个程序来询问用户是否要退出。这是我的代码：

#include<stdio.h>
#include<stdlib.h>

int ask_to_exit()
{
    int choice;
    while(choice!='y'||choice!='n')
    {   printf("Do you want to continue?(y/n):");
        while((fgetc(stdin)) != '\n');
        choice = fgetc(stdin);
        if(choice=='y') return 0;
        else if(choice=='n') return 1;
        else printf("Invalid choice!\n");
    }
}

int main()
{
    int exit = 0;
    while(!exit)
    {
        exit = ask_to_exit();
    }
    return 0;
}

由于fflush（stdin）是未定义的行为，我没有使用它。在另一个问题中遵循解决方案后，我仍然得到错误。以下是上述计划的试运行：

$./a.out
Do you want to continue?(y/n):y<pressed enter key>
<pressed enter key>
Invalid choice!
Do you want to continue?(y/n):y<pressed enter key>
<pressed enter key>
Invalid choice!
Do you want to continue?(y/n):n<pressed enter key>
<pressed enter key>
Invalid choice!
Do you want to continue?(y/n):n<pressed enter key>
n<pressed enter key>
<program exits>

Answer 1

在检查choice的值之前，您需要获取一些输入，否则它将被取消初始化。而且，在抓取第一个字符之前，你正在消耗剩余的输入，你应该在以下之后完成：

int ask_to_exit()
{
    int choice;

    do
    {
        printf("Do you want to continue?(y/n):");
        choice = fgetc(stdin);

        while (fgetc(stdin) != '\n');

        if (choice == 'y') {
            return 0;
        } else if (choice == 'n') {
            return 1;
        } else {
            printf("Invalid choice!\n");
        }
    } while (1);
}

输出中：

Do you want to continue?(y/n):g
Invalid choice!
Do you want to continue?(y/n):h
Invalid choice!
Do you want to continue?(y/n):j
Invalid choice!
Do you want to continue?(y/n):k
Invalid choice!
Do you want to continue?(y/n):m
Invalid choice!
Do you want to continue?(y/n):y
Do you want to continue?(y/n):y
Do you want to continue?(y/n):y
Do you want to continue?(y/n):n
Press any key to continue . . .

Answer 2

我在您的代码中看到以下问题。

您在初始化之前使用choice。
在阅读choice。

@WhozCraig发现错误：

choice!='y'||choice!='n'永远是真的。

我建议：

int ask_to_exit()
{
   int choice;
   while ( 1 )
   {
      printf("Do you want to continue?(y/n):");
      choice = fgetc(stdin);
      if ( choice == 'y' )
      {
         return 0;
      }
      if ( choice == 'n' )
      {
         return 1;
      }

      printf("Invalid choice!\n");

      // Skip rest of the line.
      int c;`
      while( (c = fgetc(stdin)) != EOF && c != '\n');

      // If EOF is reached, return 0 also.
      if ( c == EOF )
      {
         return 0;
      }
   }
}

使用fgetc（stdin）时的未定义行为

2 个答案: