接口实现与私有继承之间的交互

Question

When a class is inherited privately, all members of that class become private in the child class. However, in the following example, we are able to indirectly access the privately inherited doWork function implementation through AD (pointed to using a pointer to R type).

How is this allowed? Does virtual lookup ignore visibility rules since it is done at runtime?

#include <iostream>
using std::cout;

class R
{
public:
    virtual void doWork() = 0;
};

class RA : public virtual R
{
public:
    void doWork() { cout << "RA doWork\n"; };
};

class P : public virtual R, private RA
{
public:
    P() : RA() {};
};


class AD : public virtual R, private P
{
public:
    AD() : P() {};

    void doWork(int k) { cout << "AD Time dowork " << k << "\n";}
};

int main()
{
    AD ad;
    R* p = &ad;
    p->doWork();
}

The above code will print "RA doWork" when run. My expectation was that it would result in a runtime error, because doWork's definition would not be accessible from p due to the private inheritance.

Answer 1

AD inherits privately from P but publicly from R . What this means in practice is that public methods declared in R will still be accessible in AD but public methods declared in P but not in R will not be accessible in AD . So for example if P looked like:

class P : public virtual R, private RA
{
public:
    P() : RA() {};
    void doWorkP() { std::cout << "doWorkP" << std::endl; }
};

You would be able to access doWork through AD but not doWorkP .

If you want to make R 's methods inaccessible in AD simply inherit from R privately.