c++ inheritance/multi-inheritance 模棱两可的调用

Question

初学者 C++ 问题：

我有一个 class 人保护变量 FirstName 和 LastName:

class Person
{
protected:
    int Id;
    std::string FirstName;
    std::string LastName;

public:
    Person();
    Person(std::string firstName, std::string lastName);
    ~Person();
    std::string GetPersonInfo() const;
    std::string GetFirstName() const;
    std::string GetLastName() const;
};

inline std::string Person::GetPersonInfo() const {
    return FirstName + " " + LastName;
}

inline std::string Person::GetFirstName() const {
    return FirstName;
}

inline std::string Person::GetLastName() const {
    return LastName;
}

我有一个继承自 Person 的 Teacher class（还有一个 Adult Class）

class Teacher :
    public Person, public Adult
{
private:
    int ClassroomID;
public:
    Teacher() = default;
    ~Teacher() = default;
    Teacher(std::string firstName, std::string lastName, std::string emailAddress, std::string phoneNumber,
        std::vector<Address> teacherAddress, int classroomID);

};

在我的 main() 中，我有以下内容：

vector<Teacher> teachers = TeacherRepository.RetrieveTeachers();
            for (Teacher teacher : teachers) {
                cout << teacher.GetFirstName(); }

当我开始输入 "teacher." 时，我看到 "GetFirstName" 作为一个选项出现；它会引发 "Teacher::GetFirstName is ambiguous"

的编译器错误

我做错了什么？

编辑：成人定义

class Adult :
    public Person
{
protected:
    std::string Email;
    std::string PhoneNumber;
    std::vector<Address> address;
public:
    Adult() = default;
    ~Adult() = default;
    Adult(std::string emailAddress, std::string phoneNumber, std::vector<Address> address);
    Adult(std::string emailAddress, std::string phoneNumber);

};

Answer 1

您的层次结构不正确。 Teacher 同时继承了 Person 和 Adult，Adult 也继承了 Person。当你写 Teacher::GetFirstName 时，你希望编译器调用什么？也许 Person::GetFirstName 或 Adult::Person::GetFirstName。此外，您将有两个 Person 变量的示例。

决定：

虚拟继承：

class Adult : virtual public Person {...}; class Teacher : virtual public Person, public Adult {...};

更多here

老师的基本class必须是仅限成人：

class Teacher : public Adult {...};

作为不好的选项：你可以明确指出你想调用哪个方法：

`Teacher t = ...;
t.Adult::GetFirstName();`

奖励：不要按值传递参数，在您的情况下，将参数作为常量引用传递会更好。

`Person(const std::string& firstName, const std::string& lastName);`

相反

 `Person(std::string firstName, std::string lastName);`

Answer 2

您之前的设计存在多个问题。导致编译错误的直接问题是你的 Teacher 是从 Adult 和 Person class 继承的。由于 Adult 也是 Person 的子 class，因此这两个 class 都有 GetLastName() 方法，因此编译器无法判断调用哪一个。

更重要的是，您应该以更正确和清晰的方式管理您的层次结构。从语义上讲，成人也应该是人，老师也应该是成人。那么，为什么不仅从 Adult 继承 Teacher，并使您的层次结构成为设计中的线性列表？

此外，如果您想要基本 classes 的完整信息，您应该首先在派生 class 对象中初始化基本 class 对象。使用 C++ initialization list 来完成这个。

class Person
{
protected:
    int Id;
    std::string FirstName;
    std::string LastName;

public:
    Person() = default;
    Person(std::string firstName, std::string lastName) : FirstName(firstName), LastName(lastName) {}
    ~Person() = default;
    std::string GetPersonInfo() const;
    std::string GetFirstName() const;
    std::string GetLastName() const;
};

inline std::string Person::GetPersonInfo() const {
    return FirstName + " " + LastName;
}

inline std::string Person::GetFirstName() const {
    return FirstName;
}

inline std::string Person::GetLastName() const {
    return LastName;
}

class Adult : public Person
{
protected:
    std::string Email;
    std::string PhoneNumber;
    std::vector<std::string> Address;
public:
    Adult() = default;
    ~Adult() = default;
    Adult(std::string firstName, std::string lastName, std::string emailAddress, std::string phoneNumber, std::vector<std::string> address) 
        : Person(firstName, lastName), Email(emailAddress), PhoneNumber(phoneNumber), Address(address) {};
    Adult(std::string firstName, std::string lastName, std::string emailAddress, std::string phoneNumber) 
        : Person(firstName, lastName), Email(emailAddress), PhoneNumber(phoneNumber) {};

};


class Teacher : public Adult
{
private:
    int ClassroomID;
public:
    Teacher() = default;
    ~Teacher() = default;
    Teacher(std::string firstName, std::string lastName, std::string emailAddress, std::string phoneNumber,
        std::vector<std::string> teacherAddress, int classroomID) 
        : Adult(firstName, lastName, emailAddress, phoneNumber, teacherAddress), ClassroomID(classroomID) {}

};



int main(void) {

    Teacher teacher("ExampleFirstName", "ExampleLastName", "example@email.com", "100-1001000", {"example address"}, 1);

    cout << teacher.GetLastName() << endl;

    return 0;
}

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