调用基 class 的受保护函数时出错

Error while calling protected function of base class

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 class 调用基 class 的成员函数时,出现错误 statement cannot resolve address of overloaded function。我应用了作用域运算符,并且基 class 的成员函数受到保护,因此派生函数访问这些函数没有问题。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 class 。我只是在这里调用派生的 class 的函数。

C++ 允许引用成员只是命名它们。您要么需要像 &Employee::GetDes 这样的地址,要么需要像 Employee::GetDes() 这样称呼他们。您可能还想确保您确实返回了结果,但仅调用该成员就可以了,尽管它不太可能产生所需的结果。

如前所述,派生的 class Work::GetNamevoid,因此它不返回任何值。您可以将其更改为:

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

或者,如果您只想重用基础 class Employee::GetName() 但在派生 class 中将其设为 public,您需要做的就是重新在 Work.

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


[编辑] 更改了 "re-declare" 代码以遵循当前的 C++ 最佳实践,感谢@AlanStokes 的评论。具体来说,"access declarations"像原来用的那个

    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 "using declarations"。

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