c - 我该如何修复 C 中的 if 语句以使其按预期工作？

Question

在这段“联系人管理系统”的代码中，我很难获得一行的预期输出。基本上，在这一部分中，当您添加新联系人时，它会要求您“请输入公寓#”，如下所示:

 if (yes() == 1)
 {
     printf("Please enter the contact's apartment number: ");
     address->apartmentNumber = getInt();
     if (address->apartmentNumber > 0)
     {
     }
     else
     {
        printf("*** INVALID INTEGER *** <Please enter an integer>: ");
        address->apartmentNumber = getInt();
    }
  }
  else
  {
      address->apartmentNumber = 0;
  }

现在，根据我的作业，您应该输入单词(而不是数字，明白吗？)“bison”，它会显示输出:

* INVALID INTEGER * Please enter an integer:

就上下文而言，这部分工作得非常好。但是，您会被指示输入整数“-1200”，然后会出现提示

* INVALID APARTMENT NUMBER * Please enter a positive number:

这部分是我遇到问题的，因为简单地说，我不知道把它放在哪里，无论是在 if 语句中还是在 if 语句之外。我不确定，并且希望得到一些帮助。

我试图自己纠正这个问题，但它只是给了我双倍的无效整数输出，而不是这个正确的无效公寓号语句。这是我的(失败的)尝试:

    if (yes() == 1)
    {
        printf("Please enter the contact's apartment number: ");
        address->apartmentNumber = getInt();

        if (address->apartmentNumber > 0)
        {
        }
        else
        {
            printf("*** INVALID INTEGER *** <Please enter an integer>: ");
            address->apartmentNumber = getInt();
        }
        if (address->apartmentNumber < 0)
        {
        }
        else
        {
            printf("*** INVALID APARTMENT NUMBER *** <Please enter a positive number>: ");
            address->apartmentNumber = getInt();       
        }
        else
        {
            address->apartmentNumber = 0;
        }

编辑:对于那些要求 getInt() 和 yes() 代码的人，这里:

getInt()

int getInt(void)
{
    int num;
    char nl;

    scanf("%d%c", &num, &nl);
    while (nl != '\n') {
        clearKeyboard();

        printf("*** INVALID INTEGER *** <Please enter an integer>: ");
        scanf("%d%c", &num, &nl);
    }
    return num;
}

和 yes():

int yes(void)
{
    int yesno, flag;
    char c, nl;
    scanf("%c%c", &c, &nl);

    do {
        if (nl != '\n') {
            clearKeyboard();

            printf("*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: ");
            flag = 1;
            scanf("%c%c", &c, &nl);
        }
        else if (c != 'Y' && c != 'y' && c != 'N' && c != 'n') {
            printf("*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: ");
            flag = 1;
            scanf("%c%c", &c, &nl);
        }
        else if (nl == '\n' && (c == 'Y' || c == 'y' || c == 'N' || c == 'n')) 
        {
            flag = 0;
        }
    } while (flag == 1);

    if (c == 'Y' || c == 'y') {
        yesno = 1;
    }
    else {
        yesno = 0;
    }
    return yesno;
}

Answer 1

getInt 将处理非整数(文本和单词)输入，并重新提示用户输入整数，直到用户输入为止。

所以你的代码需要更像这样:

if (yes() == 1)
{
     int validNumber = 0;
     while (validNumber == 0)
     {
         printf("Please enter the contact's apartment number: ");
         address->apartmentNumber = getInt();
         if (address->apartmentNumber > 0)
         {
             validNumber = 1;
         }
         else
         {
             printf("* INVALID APARTMENT NUMBER * Please enter a positive number:\n");
         }
     }
}