Implementing multiple interfaces in c++

Question

I have the interface hierarchy as follows:

class A
{
public:
 void foo() = 0;
};

class B: public A
{
public:
void testB() = 0;
};

class C: public A
{
public:
void testC() = 0;
};

Now, I want to implement these interfaces by the same hierarchy, that is a Base class AImpl, BImpl and CImpl but I am not sure how to derive them from their corresponding interfaces.

Please help. Thanks in advance.

Answer 1

You can implement each individial interface using a separate template and then chain the templates to construct the derived object as if from building blocks. This method was also used by venerable ATL library to implement COM interfaces (for those of us old enough).

Note that you don't need virtual inheritance for that.

I slightly modified you example for a more complex derivation C -> B -> A to show how this method scales easily:

#include <stdio.h>

// Interfaces

struct A
{
    virtual void foo() = 0;
};

struct B : A
{
    virtual void testB() = 0;
};

struct C : B
{
    virtual void testC() = 0;
};

// Implementations

template<class I>
struct AImpl : I
{
    void foo() { printf("%s\n", __PRETTY_FUNCTION__); }
};

template<class I>
struct BImpl : I
{
    void testB() { printf("%s\n", __PRETTY_FUNCTION__); }
};

template<class I>
struct CImpl : I
{
    void testC() { printf("%s\n", __PRETTY_FUNCTION__); }
};


// Usage

int main() {
    // Compose derived objects from templates as from building blocks.
    AImpl<A> a;
    BImpl<AImpl<B> > b;
    CImpl<BImpl<AImpl<C> > > c;

    a.foo();

    b.foo();
    b.testB();

    c.foo();
    c.testB();
    c.testC();
}

Output:

void AImpl<I>::foo() [with I = A]
void AImpl<I>::foo() [with I = B]
void BImpl<I>::testB() [with I = AImpl<B>]
void AImpl<I>::foo() [with I = C]
void BImpl<I>::testB() [with I = AImpl<C>]
void CImpl<I>::testC() [with I = BImpl<AImpl<C> >]