Question

我正在尝试获取函数membername以返回字符串staffmember，而usermember使用数据类型字符串是不可能的吗？如果没有，我可以使用其他数据类型吗？

    string membername();
    {
        if (member == 1)
        {
            string studentmember;
            cout << "Enter Student Member Name (possible names: Tom, Max, Ben): ";
            cin >> studentmember;
            return studentmember;
        }
        if (member == 2)
        {
            string staffmember;
            cout << "Enter Staff Member Name (possible names: Linda, Mary, Bob): ";
            cin >> staffmember;
            return staffmember;
        }
        else
        {
            cout << "Enter a valid number" << endl;
        }
    }

Answer 1

您可以执行以下操作：

struct Person
{
    int id;  // Every person has an ID.
    std::string name;  // Every person has a name.

    // Every Person has a function to input their name,
    //    implemented (specialized) by the child.
    virtual void    get_name() = 0;
};

struct Student : public Person // A student is-a person
{
    void get_name()
    {
        std::cout << "Enter Student's name:";
        std::getline(std::cin, name);
    }
};

struct Staff : public Person // A staff member is-a person
{
    void get_name()
    {
        std::cout << "Enter Staff's name:";
        std::getline(std::cin, name);
    }
};

学生对象和职员对象具有不同的get_name方法。

您可以将它们作为一个人来对待：

Person * p1 = new Student;
Person * p2 = new Staff;
p1->get_name(); // Get the name of the person, using appropriate prompt.
p2->get_name(); // Uses the Staff::get_name() function.

对于模板版本，您必须使用用于人员类型的方法（或成员）。

template <typename College_Person>
void get_name(College_Person& cp)
{
   const std::string person_type_name = cp.get_type_name();
   std::cout << "Enter " << person_type_name << " name: ";
   std::getline(std::cin, cp.name);
}

或

template <typename College_Person>
void get_name(College_Person& cp)
{
    std::cout << "Enter "
              << cp.person_type_name
              << " name: ";
    std::getline(std::cin, cp.name);
}

C ++从到不存在合适的转换函数

1 个答案: