#include <stdio.h>
   void clearKeyboard(void)
   {
       while (getchar() != '\n')   ; // empty execution code block on purpose
   }
   int yes(void)
   {

       char a,b;
       printf("<Please enter a character>: ");
       scanf("%c%c", &a,&b);

       while ((a !='Y' && a !='y' && a !='N' && a!='n') || (b!='\n'))
       {
          if (b!='\n') ungetc(b, stdin),scanf("%*[^\n]%c", &b);
          a='n',b='\n';
          clearKeyboard();
          printf("*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: ");
          scanf("%c%c", &a,&b);

        }
        if (a =='Y' || a=='y')
        {
            printf("Contact Management System: terminated\n");
            return 1;

        }
        else
        {
            if (a =='N' || a=='n')
            ContactManagerSystem();
            return 0;
        }

   }

   int menu(void)

   {
       int i;
       printf("0. Exit\n\nSelect an option:> ");
       scanf("%d", &i);
       while (i<0 || i>6)
       {
           printf("*** OUT OF RANGE *** <Enter a number between 0 and 6>: ");
           scanf("%d", &i);
       }
       return i;
   }

   void ContactManagerSystem(void)

   {
     int i=menu();

      switch(i)
      {
         case 0 :
          printf("Exit the program? (Y)es/(N)o: ");
          yes();


       }

   }
   int main(void)
   {
       ContactManagerSystem();
   }







So my "yes" function is working fine on it's own but when I call it within


ContactManagerSystem()case 0由于某种原因而变得疯狂
    “是”功能实际上是一个验证检查功能,我需要用户
    输入“是”或“否”,如“ Y”,“ y” N”或“ n”
    等待用户输入“ Y”,“ y”,“ N”或“ n”(是或否)。
    如果用户回答是(“ Y”,“ y”),它将结束显示
    以下消息:


  联系人管理系统:已终止      否则,如果用户输入否(“ N”,“ n”),则应用程序继续执行
      显示菜单。

最佳答案

我认为您需要重新考虑对&&||的使用,并使用括号。这段代码还照顾到EOF和键入的单词(如“是”)。

#include <stdio.h>

int main(void)
{
    char a, b;
    printf("<Please enter a character>: ");
    if (scanf("%c%c", &a, &b) == 2)
    {
        while ((a != 'Y' && a != 'y' && a != 'N' && a != 'n') || b != '\n')
        {
            printf("*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: ");
            if (b != '\n')
            {
                int c;
                while ((c = getchar()) != EOF && c != '\n')
                    ;
            }
            if (scanf("%c%c", &a, &b) != 2)
                break;
        }
        if (a == 'Y' || a == 'y')
        {
            printf("Got yes!\n");
            return 1;
        }
        else if (a == 'N' || a == 'n')
        {
            printf("Got no!\n");
            return 0;
        }
    }
    printf("Got EOF\n");
    return 0;
}


如果a中的字母既不是'Y'也不是'y'也不是'N'也不是'n',或者b中的值不是换行符,则报告问题,请在所有换行符前添加所有杂散字符(或EOF),然后继续。通过打印报告状态以及退出状态。

示例运行:

$ ./scan13; echo $?
<Please enter a character>: yes
*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: no
*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: pdq
*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: y
Got yes!
1
$ scan13; echo $?
<Please enter a character>: n
Got no!
0
$ ./scan13; echo $?
<Please enter a character>: Got EOF
0
$


如以下注释中所述,scanf()的问题之一是确保错误行为符合预期。通常更容易使用fgets()或POSIX getline()将整行读入缓冲区,然后使用sscanf()解析数据。

关于c - 函数获取垃圾值,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/47804105/

10-16 23:34