在派生中创建构造函数时 class

Question

When creating a constructor in a derived class, Using the base class's private member variable, Can it be given as an initial value for a derived class's constructor parameter?

class Base{
    private: int m_Par;
    protected: int m_Pro;
    public:int m_Pub;
};
class Derived :public Base{
    public:Derived(int m_Par=1,int m_Pro=2,int m_Pub=3);
           void Func();
};
Derived::Derived(int m_Par,int m_Pro,int m_Pub)
{
    //this->m_Par=m_Par;
    this->m_Pro=m_Pro;
    this->m_Pub=m_Pub;
}

I remember that derived classes can't access the private member variables of the base class, However, I wonder why there is no error when the private member variable of the base function is given as the initial value of the derived class parameter.

Answer 1

are you compiling with the line commented out? because when i compile with g++ the compiler complaints with the error below.

derived.cpp: In constructor ‘Derived::Derived(int, int, int)’:
derived.cpp:12:11: error: ‘int Base::m_Par’ is private within this context
this->m_Par=m_Par;
       ^~~~~
derived.cpp:2:18: note: declared private here
private: int m_Par;

Answer 2

Parameter is not the field. When you write parameter names to your derived classes constructor like Derived(int m_Par=1,int m_Pro=2,int m_Pub=3) it doesn't match with the fields. You are matching fields with the parameters in the function body if you want so.

If you want to initialize a private field in base class you can use something like that.

class Base{
    private: int m_Par;
    protected: int m_Pro;
    public:int m_Pub;

    Base(int m_Par){
        this->m_Par = m_Par;
    }

    int getM_Par(){
        return this->m_Par;
    }
};
class Derived :public Base{
    public:Derived(int m_Par=1,int m_Pro=2,int m_Pub=3);
           void Func();
};
Derived::Derived(int m_Par,int m_Pro,int m_Pub) : Base(m_Par)
{
    this->m_Pro=m_Pro;
    this->m_Pub=m_Pub;
}

int main(){
   Derived d(10,5,3);
   std::cout<<"M_Par value: "<<d.getM_Par();
   return 0;
}

EDIT: In the code below you wrote this->m_Pro = m_Pro; and this->m_Pub=m_Pub; . Those code blocks assigns the values taken from parameters to your base classes fields. When you write Derived:: Derived(int m_Par =1, int m_Pro=2, int m_Pub=3) what you are doing is that if user doesn't pass any parameters to the function your parameters will have default values that you have assigned. So parameters and the fields are not the same. The fields you wrote in the base class are belong to base class but parameters are local variables belong to the function you passed into.

Derived::Derived(int m_Par,int m_Pro,int m_Pub) : Base(m_Par)
{
    this->m_Pro=m_Pro;
    this->m_Pub=m_Pub;
}

Answer 3

First of all, m_Par is the local variable in your constructor and has nothing to do with the member m_Par of the Base class. In order to access the member with the same name as the local variable use this->m_Par . However, You can't access private members of the Base class in the constructor of your Derived class.

I think that you just complicated your question when mixed naming convention for local variables and class members . Prefix m_ is usually added to the names of class members:

Base(int par){
    this->m_Par = par;
}

Derived::Derived(int par,int pro,int pub) : Base(par)
{
    this->m_Pro = pro;
    this->m_Pub = pub;
}