C++:链表、字符串指针

Question

我在实现链表以及将列表的值返回到屏幕时遇到问题。如果之前有人问过这个问题，请原谅 - 我找不到它，也许我不知道如何正确地提出这个问题!

我只是在构建它以获得更多功能之前测试这段代码，但我似乎无法开始。我在这里注释掉了一些代码，以备将来使用。我没有包括下面的其他方法，只有 Display，特别是因为它们什么都没有。

我运行了 cout 语句以查看我在以下语句中遇到了问题:person->lname。这将编译，但不会运行。当我尝试调试时，出现字符串 header 错误。

非常感谢任何帮助!

      #include <iostream>
        #include <string>
        using namespace std;

    struct Birthday
    {
        int month;
        int day;
        int year;
    };

    struct Anniversary
    {
        int month;
        int day;
        int year;
    };

    struct Address
    {
        string street;
        string city;
        string state;
        int zip;
    };

    struct PeopleInfo
    {
        string lname;
        string fname;
        Address fulladdr;
        Birthday bday;
        Anniversary aday;
        //PeopleInfo* nextperson;
    };

    class AddressBook
    {
    public:
        AddressBook();
        AddressBook(string);
        void NewPerson();
        void DeletePerson();
        void SetName();
        void SetAddress();
        void SetDates();
        void Search(string);
        void Sort();
        void Display();
        void BDayCard();
        void ADayCard();
    private:
        PeopleInfo* person;
        //PeopleInfo* currentperson; 
        int length;
    };

    AddressBook::AddressBook()
    {
        person->lname = "last";
        person->fname = "first";
        person->fulladdr.street = "default st.";
        person->fulladdr.city = "anytown";
        person->fulladdr.state = "NJ";
        person->fulladdr.zip = 00000;
        person->bday.month = 0;
        person->bday.day = 0;
        person->bday.year = 0;
        person->aday.month = 0;
        person->aday.day = 0;
        person->aday.year = 0;
        length = 0;
        //person->nextperson->lname = NULL;

        //currentperson->nextperson->lname = NULL;
    }
    void AddressBook::Display()
    {
        cout << person->lname << ", " << person->fname << " " << endl
            << person->fulladdr.street << endl
            << person->fulladdr.city << ", " << person->fulladdr.state << " " << person->fulladdr.zip << endl
            << "Birthday: " << person->bday.month << "/" << person->bday.day << "/" << person->bday.year << endl
            << "Anniversary: " << person->aday.month << "/" << person->aday.day << "/" << person->aday.year << endl << endl;
    }

// Function Prototypes
void Menu(char &entry);
void Action(char &entry, AddressBook AllNames);

// Main
int main()
{
    cout << "test\n";
    char entry;
    cout << "test\n";
    AddressBook AllNames;
    cout << "test\n";
    do
    {
        Menu(entry);
        Action(entry, AllNames);
    } while (!(entry == 'E' || entry == 'e'));
return 0;
}

void Menu(char &entry)
{
    cout << "\nADDRESSBOOK\n\n"
        << "(N) Enter New Name\n"
        << "(D) Delete a Name\n"
        << "(C) Change a Name/Date\n"
        << "(U) Update Anniversary/Birthday\n"
        << "(S) Show Address Book Entries\n"
        << "(B) Make a Birthday Card\n"
        << "(A) Make an Anniversary Card\n"
        << "(E) Exit Program\n\n";
    cin >> entry;
    cout << "\n";
}

void Action(char &entry, AddressBook AllNames)
{
    switch (entry)
    {
    case 'N':
    case 'n': AllNames.NewPerson(); break;
    case 'D':
    case 'd': AllNames.DeletePerson(); break;
    case 'C':
    case 'c': AllNames.SetName(); break;
    case 'U':
    case 'u': AllNames.SetDates(); break;
    case 'S':
    case 's': AllNames.Display(); break;
    case 'B':
    case 'b': AllNames.BDayCard(); break;
    case 'A':
    case 'a': AllNames.ADayCard(); break;
    case 'E':
    case 'e': break;
    default: "\nPlease input a valid entry\n";      
    }
}

Answer 1

你需要初始化“person”。

您在“AllNames”类中定义了“PeopleInfo* person”，这意味着 person 是指向“PeopleInfo”类型内存的指针。

问题是它实际上并没有指向任何东西。你只是说它可以指向这样一个类。

由于“person”为 null 或垃圾(您什么都没做)，尝试将某些内容分配给 person->lname 将导致程序崩溃。如果您尝试调试 person->lname 的内容，那也会崩溃。

解决问题:

在 AllNames 构造函数的第一行之前(右边 before "person->lname = "last";")放置命令:

person = new PeopleInfo();