Which function will it call?

Here is the code, I wrote the comments. The question is I don't know which function will be called after the function unhide in the Derive class.

    #include <CONIO.H>
    #include <IOSTREAM>
    #include <string>

    using namespace std;

    class Base
    {
        string strName;
    public:
        Base& operator=(const Base &b)
        {
            this->strName = b.strName;
            cout << "copy assignment" << endl;
            return *this;
        }
        Base& operator=(string& str)
        {
            this->strName = str;
            cout << "operator=(string& str)" << endl;
            return *this;
        }

    };

    class Derive : public Base
    {
    public:
        int num;
        using Base::operator =; // unhide Base::operator=();
    };

    int main(int argc, char *argv[])
    {
        Derive derive1;
        derive1.num = 1;

        Derive derive2;

        Base b1;
        derive1 = b1;  // This will call Base& Base::operator=(const Base &b)
                           //no problem

        string str("test");
        derive1 = str;  // This will call Base& Base::operator=(string& str)
                            // no problem

        derive2 = derive1; // What function will this statement call???
                               // If it calls Base& Base::operator(const Base &b)
                               // how could it be assigend to a class Derive?
        return 0;
    }

But the result of the code is: derive2.num equals to 1!!!, that means the whole class has been copied after the statment, why would this happen?

Thanks to Tony, I think I got the answer.

here is my explanation:

Based on C++0x 7.3.3.3 and 12.8.10, The using-statement in Derive will be explained like this

class Derive : public Base
{
public:
    int num;
    //using Base::operator =;
    Base& operator=(const Base &b); // comes form the using-statement
    Base& operator=(string& str); // comes form the using-statement
    Derive& operator=(const Derive &); // implicitly declared by complier
};

So when I wrote:

string str("test");
derive1 = str;

function Base& Base::operator=(string& str); will be called,

and when I wrote:

Base b1;
derive1 = b1;

function Base& Base::operator=(const Base &b); will be called,

开发者_运维技巧

finnaly, when I wrote:

derive2 = derive1;

function Derive& Dervie::operator=(const Derive&); will be called.

---

Standard 7.3.3-4 (from an old draft, but in this regard still valid):

If an assignment operator brought from a base class into a derived class scope has the signature of a copy-assignment operator for the derived class (class.copy), the using-declaration does not by itself suppress the implicit declaration of the derived class copy-assignment operator; the copy-assignment operator from the base class is hidden or overridden by the implicitly-declared copy-assignment operator of the derived class, as described below.

So, the implicit Derived::operator=() is used.

---

It will call the derived operator=, which in its automatically generated implementation will call operator= from Base as well as copy the members in Derive.