这是我的yes（）代码的样子。我试图让它成为如果用户输入除了＆＃39; y＆＃39;之外的任何东西。 ＆＃39; Y＆＃39; ＆＃39; N＆＃39; ＆＃39; N＆＃39; 。我解决了输入除了那些字母之外的任何问题的问题，但是如果我要输入Yes或No，它只需要第一个字符，它使函数认为是其中一个字母。 我不知道如何解决这个问题。
int yes(void) {
char singleLetter = ' ';
int finalValue = -1;
int theResult = 0;
scanf(" %c", &singleLetter);
clearKeyboard();
do
{
switch (singleLetter)
{
case 'Y':
case 'y':
finalValue = 1;
theResult = 1;
break;
case 'N':
case 'n':
finalValue = 0;
theResult = 1;
break;
default:
theResult = 0;
printf("Only (Y)es or (N)o are acceptable: ");
scanf("%c", &singleLetter);
clearKeyboard();
}
} while (!theResult);
return finalValue;
}
我遇到的第二个问题是我得到了这个陈述：
此功能的目的是设置联系人使用的值 指针参数变量（设置它指向的联系人）。
使用接收到此功能的指针参数来提供 适当的联系成员到“get”函数（getName， getAddress和getNumbers）设置联系人的值。
目前这是＆＃34; get＆＃34;部分：
void getName(struct Name *contactName) {
printf("Please enter the contact's first name: ");
scanf("%s", (*contactName).firstName);
printf("Do you want to enter a middle intial(s)? (y or n): ");
if (yes() == 1) {
printf("Please enter the contact's middle intial(s): ");
scanf("%s", (*contactName).middleInitial);
}
printf("Please enter the contact's last name: ");
scanf("%s", (*contactName).lastName);
}
// getAddress:
void getAddress(struct Address *
contactAddress) {
printf("Please enter the contact's street number: ");
(*contactAddress).streetNumber == getInt();
printf("Please enter the contact's street name: ");
scanf(" %[^\n]", (*contactAddress).street);
printf("Do you want to enter an apartment number? (y or n): ");
if (yes() == 1) {
printf("Please enter the contact's apartment number: ");
scanf("%d", (*contactAddress).apartmentNumber);
}
printf("Please enter the contact's postal code: ");
scanf(" %[^\n]", (*contactAddress).postalCode);
printf("Please enter the contact's city: ");
scanf("%s", (*contactAddress).city);
}
// getNumbers:
// getNumbers:
// NOTE: Also modify this function so the cell number is
// mandatory (don't ask to enter the cell number)
void getNumbers(struct Numbers *contactNumber) {
printf("Please enter the contact's cell phone number: ");
scanf(" %s", (*contactNumber).cell);
printf("Do you want to enter a home phone number? (y or n) ");
if (yes() == 1) {
printf("Please enter the contact's home phone number: ");
scanf("%s", (*contactNumber).home);
}
printf("Do you want to enter a business number? (y or n) ");
if (yes() == 1) {
printf("Please enter the contact's business phone number: ");
scanf("%s", (*contactNumber).business);
}
printf("\n");
}
然后我在下面有这个。而且我不确定该部分的编程方式。我试着把正确的东西（但显然是错的）
void getContact(struct Contact *contact) {
getName(contact);
getAddress(contact);
getNumbers(contact);
}
这些是我教授所宣称的（所以我不能改变这一部分）：
getContact(&contact);
printf("\nValues Entered:\n");
printf("Name: %s %s %s\n", contact.name.firstName, contact.name.middleInitial, contact.name.lastName);
printf("Address: %d|%s|%d|%s|%s\n", contact.address.streetNumber, contact.address.street,
contact.address.apartmentNumber, contact.address.postalCode, contact.address.city);
printf("Numbers: %s|%s|%s\n", contact.numbers.cell, contact.numbers.home, contact.numbers.business);
这是我头文件中的内容：
struct Name {
char firstName[31];
char middleInitial[7];
char lastName[36];
};
// Structure type Address declaration
// Place your code here...
struct Address {
char street[41];
int streetNumber[1];
int apartmentNumber[1];
char postalCode[8];
char city[41];
};
// Structure type Numbers declaration
// Place your code here...
struct Numbers {
char cell[21];
char home[21];
char business[21];
};
// Structure type Contact declaration
// Place your code here...
struct Contact {
struct Name name;
struct Address address;
struct Numbers numbers;
};
//------------------------------------------------------
// Function Prototypes
//------------------------------------------------------
// +-------------------------------------------------+
// | NOTE: Copy/Paste your Assignment-2 Milestone-1 |
// | function prototypes here... |
// +-------------------------------------------------+
// Get and store from standard input the values for Name
// Place your code here...
void getName(struct Name *);
// Get and store from standard input the values for Address
// Place your code here...
void getAddress(struct Address *);
// Get and store from standard input the values for Numbers
// Place your code here...
void getNumbers(struct Numbers *);
// Get and store from standard input the values for a Contact
// Place your code here...
void getContact(struct Contact *);
使用以下说明运行程序时的结果
------------------------------------------
Testing: getContact(struct Contact *)
------------------------------------------
Please enter the contact's first name: Andrew
Do you want to enter a middle intial(s)? (y or n): n
Please enter the contact's last name: Random
Please enter the contact's street number: 100
Please enter the contact's street name: Rain
Do you want to enter an apartment number? (y or n): y
Please enter the contact's apartment number: 14
Please enter the contact's postal code: Z8Z 7Q7
Please enter the contact's city: Toronto
Please enter the contact's cell phone number: 647-999-9999
Do you want to enter a home phone number? (y or n) n
Do you want to enter a business number? (y or n) n
Values Entered:
Name: 647-999-9999 Random
Address: 13630948||13630952||
Numbers: ||
正如您所看到的，我的名字是我的手机号码，姓氏有效，地址根本不起作用，而且数字不存在。
答案 0 :(得分：0)
我认为主要的问题是，在getContact（）函数中，您将指针传递给Contact结构，而函数getName（）等则需要指向内部结构（例如Name）的指针。将其更改为：
void getContact(struct Contact *contact) {
getName(&contact->name);
getAddress(&contact->address);
getNumbers(&contact->numbers);
}
请开始阅读警告信息！