Refactoring a leaf class to a base class, and keeping it also a interface implementation

Question

I am trying to refactor a working code. The code basically derives an interface class into a working implementation, and I want to use this implementation outside the original project as a standalone class.

However, I do not want to create a fork, and I want the original project to be able to take out their implementation, and use mine. The problem is that the hierarchy structure is very different and I am not sure if this would work. I also cannot use the original base class in my project, since in reality it's quite entangled in the project (too many classes, includes) and I need to take care of only a subdomain of the problems the original project is.

I wrote this code to test an idea how to implement this, and while it's working, I am not sure I like it:

#include <iostream>

// Original code is:
// IBase -> Derived1

// I need to refactor Derive2 to be both indipendet class
// and programmers should also be able to use the interface class
// Derived2 -> MyClass + IBase
// MyClass


class IBase {
public:
    virtual void printMsg() = 0;
};

///////////////////////////////////////////////////
class Derived1 : public IBase {
public:
    virtual void printMsg(){ std::cout << "Hello from Derived 1" << std::endl; }
};


//////////////////////////////////////////////////
class MyClass {
public:
    virtual void printMsg(){ std::cout << "Hello from MyClass" << std::endl; }
};

class Derived2: public IBase, public MyClass{
    virtual void printMsg(){ MyClass::printMsg(); }
};

class Derived3: public MyClass, public IBase{
    virtual void printMsg(){ MyClass::printMsg(); }
};

int main()
{
    IBase *o1 = new Derived1();
    IBase *o2 = new Derived2();
    IBase *o3 = new Derived3();
    MyClass *o4 = new MyClass();

    o1->printMsg();
    o2->printMsg();
    o3->printMsg();
    o4->printMsg();

    return 0;
}

The output is working as expected (tested using gcc and clang, 2 different C++ implementations so I think I am safe here):

[elcuco@pinky ~/src/googlecode/qtedit4/tools/qtsourceview/qate/tests] ./test1
Hello from Derived 1
Hello from MyClass
Hello from MyClass
Hello from MyClass
[elcuco@pinky ~/src/googlecode/qtedit4/tools/qtsourceview/qate/tests] ./test1.clang 
Hello from Derived 1
Hello from MyClass
Hello from MyClass
Hello from MyClass

The question is

My original code was:

class Derived3: public MyClass, public IBase{
    virtual void IBase::printMsg(){ MyClass::printMsg(); }
};

Which is what I want to express, but this does not compile. I must admit I do not fully understand why this code work, as I expect that the new method Derived3::printMsg() will be an implementation of MyClass::printMsg() and not IBase::printMsg() (even tough this is what I do want).

How does the compiler chooses which method to re-implement, when two "sister classes" have the same virtual function name?

If anyone has a better way of implementing this, I would like to know as well :)

Answer 1

The answer is, compiler overrides both functions, as can be shown by this sample:

#include <cstdio>
using std::printf;

class A {
public:
    virtual void a() {
        printf("A::a\n");
    }
};
class B {
public:
    virtual void a() {
        printf("B::a\n");
    }
};
class C : public A, public B {
public:
    virtual void a() {
        printf("C::a\n");
        A::a();
        B::a();
    }
};
int main() {
    C c;
    A &a = c;
    B &b = c;
    printf("Calling C::a via A\n");
    a.a();
    printf("Calling C::a via B\n");
    b.a();
}

The output is:

Calling C::a via A
C::a
A::a
B::a
Calling C::a via B
C::a
A::a
B::a

If you want to override one and not the other, you need to rename it. Not only it will do what you want, but it will be clearer as well.

Answer 2

The compiler doesn't actually choose one of the two methods to override. Instead it spots that there are two virtual functions with identical signatures and looks for an override for that signature in derived classes.

If Derived3 had not re-implemented printMsg then calling that member on an object of type Derived3 would have been ambiguous. Since you have re-implemented the method the compiler can happily and unambiguously call that version on objects of type Derived3 .