c++ - 调用基类的 protected 函数时出错

Question

我是 C++ 的初学者。我正在研究继承。在我的代码中，当我尝试从派生类调用基类的成员函数时，出现错误 statement cannot resolve address of overloaded function .我已经应用了作用域运算符并且基类的成员函数受到保护，因此派生类在访问这些函数时没有问题。我在这里做错了什么？这是我的代码:

#include <iostream>
#include <string.h>

using namespace std;
class Employee
{
private:
    char *emp_name;
    int emp_code;
    char *designation;

protected:
    Employee(char *name ="",int code = 0,char *des = "", int nlength=1, int dlength=1): emp_code(code)
    {
        emp_name = new char[nlength+1];
        strncpy(emp_name,name,nlength);

        designation = new char[dlength+1];
        strncpy(designation,des,nlength);
    }

    ~Employee()
    {
        delete[] emp_name;
        delete[] designation;
    }

    char* GetName()
    {
        return emp_name;
    }

    int GetCode()
    {
        return emp_code;
    }

    char* GetDes()
    {
        return designation;
    }
};

class Work: public Employee
{
private:
    int age;
    int year_exp;

public:
    Work(char *name ="", int code = 0, char *des = "", int nlength=1, int dlength=1, int w_age=0, int w_exp=0):Employee(name,code,des,nlength,dlength),age(w_age),year_exp(w_exp)
    {
    }

    void GetName()
    {
        Employee::GetName;
    }

    void GetCode()
    {
        Employee::GetCode;
    }

    void GetDes()
    {
        Employee::GetDes;
    }

    int GetAge()
    {
        return age;
    }

    int GetExp()
    {
        return year_exp;
    }

};

int main()
{
    using namespace std;
    Work e1("Kiran",600160,"Implementation Specialist", strlen("Kiran"),strlen("Implementation Specialist"),24,5);

    cout << "Name: " << e1.GetName() << endl;
    cout << "Code: " << e1.GetCode() << endl;
    cout << "Designation: " << e1.GetDes() << endl;
    cout << "Age: " << e1.GetAge() << endl;
    cout << "Experience: " << e1.GetExp() << endl;
}

此外，我收到错误 no-match for operator<< .我没有打印任何类(class)。我这里只是调用派生类的函数。

Answer 1

如前所述，派生类 Work::GetName是void所以它没有返回任何值。您可以将其更改为:

class Work: public Employee
{
//...
public:
//...
    char *GetName()
    {
        return Employee::GetName();
    }

或者，如果您只想重用基类 Employee::GetName()但是在派生类中将其公开，您需要做的就是在 Work 中重新将其声明为公开.

class Work: public Employee
{
//...
public:
//...
    using Employee::GetName; // <--- this re-declares the inherited GetName() as public

---

[编辑] 更改了“重新声明”代码以遵循当前的 C++ 最佳实践，感谢@AlanStokes 的评论。具体来说，像最初使用的那样的“访问声明”

    Employee::GetName; // <--- this re-declares the inherited GetName() as public

自 C++98 以来已弃用，取而代之的是具有相同效果的“使用声明”。

    using Employee::GetName; // <--- this re-declares the inherited GetName() as public