Question

Possible Duplicate:
Why does C++ not let baseclasses implement a derived class' inherited interface?

#include <iostream>

class Interface
{
public:
    virtual void yell(void) = 0;
};

class Implementation
{
public:
    void yell(void)
    {
        std::cout << "hello world!" << std::endl;
    }
};

class Test: private Implementation, public Interface
{
public:
    using Implementation::yell;
};

int main (void)
{
    Test t;
    t.yell();
}

I want the Test class to be implemented in terms of Implementation, and I want to avoid the need to write the

void Test::yell(void) { Implementation::yell(); }

method. Why it is not possible to do it this way? Is there any other way in C++03?

Answer 1

using only brings a name into a scope.

It doesn’t implement anything.

If you want Java-like get-implementation-by-inheritance, then you have to explicitly add the overhead associated with that, namely virtual inheritance, like this:

#include <iostream>

class Interface
{
public:
    virtual void yell() = 0;
};

class Implementation
    : public virtual Interface
{
public:
    void yell()
    {
        std::cout << "hello world!" << std::endl;
    }
};

class Test: private Implementation, public virtual Interface
{
public:
    using Implementation::yell;
};

int main ()
{
    Test t;
    t.yell();
}

---

EDIT: a bit sneaky this feature, I had to edit to make the code compile with g++. Which didn’t automatically recognize that the implementation yell and the interface yell were one and the same. I’m not completely sure what the standard says about that!