C 如何解决此输入缓冲问题？

我的代码有一个问题，输出将比我希望的更早地打印语句，我被告知这是由于输入缓冲，在scanf中的“%”之前添加一个空格可以解决这个问题，但这仍然是一个反复出现的问题

printf（“请输入联系人的名字：”）； scanf（“%31s”，name.firstName）

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您的示例显示了要由

scanf（“%41s”，address.street）扫描的输入是“基尔街”。但是
%s
格式说明符在第一个空格处停止，因此这不起作用。我建议
scanf（“%41[^\n]”，address.street）。您只需要将前导空格放在前面，并且
%c
，因为其他格式会自动过滤前导空格。类似地，对于可能不止一个单词的任何其他字符串，例如城市“纽约”和姓氏“Da Silva”等，您还应始终检查
scanf
的返回值，即成功扫描的项目数。回到前面的评论，我认为清除输入缓冲区不是问题，而是一个字符串中有多个单词。我认为发布了错误的副本。问题不在于清除输入缓冲区（OP已采取步骤清除前导空格），而在于不读取多个字符串的整个输入。
printf("Do you want to enter a middle initial(s)? (y or n): ");
scanf(" %c", &option);


if (option == 'y' || option == 'Y')
{
printf("Please enter the contact's middle initial(s): ");
scanf("  %7s", name.middleInitial);
}

printf("Please enter the contact's last name: ");
scanf(" %36s", name.lastName);


// Contact Address Input:

printf("Please enter the contact's street number: ");
scanf(" %d", &address.streetNumber);

printf("Please enter the contact's street name: ");
scanf(" %41s", address.street);

printf("Do you want to enter an apartment number? (y or n): ");
scanf(" %c", &option);

if (option == 'y' || option == 'Y')
{
    printf("Please enter the contact's apartment number: ");
    scanf(" %d", &address.apartmentNumber);
}

printf("Please enter the contact's postal code: ");
scanf(" %8s", address.postalCode);

printf("Please enter the contact's city: ");
scanf(" %41s", address.city);

// Contact Numbers Input:
printf("Do you want to enter a cell phone number? (y or n): ");
scanf(" %c", &option);

if (option == 'y' || option == 'Y')
{
    printf("Please enter the contact's cell phone number: ");
    scanf(" %11s", numbers.cell);
}

printf("Do you want to enter a home phone number? (y or n): ");
scanf(" %c", &option);

if (option == 'y' || option == 'Y')
{
    printf("Please enter the contact's home phone number: ");
    scanf(" %11s", numbers.home);
}

printf("Do you want to enter a business phone number? (y or n): ");
scanf(" %c", &option);

if (option == 'y' || option == 'Y')
{
    printf("Please enter the contact's business phone number: ");
    scanf(" %11s", numbers.business);
}