inheritance - C++: using a base class as the implementation of an interface

Question

In C++ is it possible to use another base class to provide the implementation of an interface (i.e. abstract base class) in a derived class?

class Base
{
    virtual void myfunction() {/*...*/};
}
class Interface
{
    virtual void myfunction() = 0;
}
class Derived
    : public Base, public Interface
{
    // myfunction is implemented by base
}

The above should compile, but doesn't actually work. Is there any way to achieve this effect?

In case anyone cares, the reason for wanting this is that it make sense (for my application) to use a generic library from another project/namespace to provide the implementation of an interface in my project. I could just wrap everything, but that seems like a lot of extra overhead.

Thanks.

Answer 1

If Base isn't derived from Interface, then you'll have to have forwarding calls in Derived. It's only "overhead" in the sense that you have to write extra code. I suspect the optimizer will make it as efficient as if your original idea had worked.

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

class Base {
    public:
        virtual void myfunction() {/*...*/}
};

class Derived : public Interface, public Base {
    public:
        void myfunction() { Base::myfunction(); }  // forwarding call
};

int main() {
   Derived d;
   d.myfunction();
   return 0;
}