c++ - 它会调用哪个函数？

Question

这是代码，我写了注释。 问题是我不知道在Derive类中函数unhide后会调用哪个函数。

    #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;
    }

但是代码的结果是:derive2.num等于1!!!，也就是说语句之后整个类都被复制了，为什么会这样呢？

多亏了托尼，我想我找到了答案。

这是我的解释:

基于C++0x 7.3.3.3和12.8.10，Derive中的using语句会这样解释

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
};

所以当我写的时候:

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

函数 Base& Base::operator=(string& str); 将被调用，

当我写的时候:

Base b1;
derive1 = b1;

函数 Base& Base::operator=(const Base &b); 将被调用，

最后，当我写道:

derive2 = derive1;

函数 Derive& Dervie::operator=(const Derive&); 将被调用。

Answer 1

标准 7.3.3-4(来自旧草案，但在这方面仍然有效):

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.

因此，使用隐式 Derived::operator=()。