How to override a function in another base class?

Question

I'm not exactly sure the terminology to use, but here's my example:

class Base {
public:
    virtual void test() = 0;
};

class Mixin {
public:
    virtual void test() { }
};

class Example : public Base, public Mixin {
};

int main(int argc, char** argv) {
    Example example;
    example.test();
    return 0;
}

I want my Mixin class to implement the pure virtual function Base::test, but when I compile this, it says:

test.cpp: In function ‘int main(int, char**)’:
test.cpp:15:13: error: cannot declare variable ‘example’ to be of abstract type ‘Example’
     Example example;
             ^
test.cpp:11:7: note:   because the following virtual functions are pure within ‘Example’:
 class Example : public Base, public Mixin {
       ^
test.cpp:3:18: note:    virtual void Base::test()
     virtual void test() = 0;
                  ^
test.cpp:16:13: error: request for member ‘test’ is ambiguous
     example.test();
             ^
test.cpp:8:18: note: candidates are: virtual void Mixin::test()
     virtual void test() { }
                  ^
test.cpp:3:18: note:                 virtual void Base::test()
     virtual void test() = 0;
                  ^

I can add a using statement to make it not ambiguous:

class Example : public Base, public Mixin {
public:
    using Mixin::test;
};

But it says I still haven't implemented it:

test.cpp: In function ‘int main(int, char**)’:
test.cpp:17:13: error: cannot declare variable ‘example’ to be of abstract type ‘Example’
     Example example;
             ^
test.cpp:11:7: note:   because the following virtual functions are pure within ‘Example’:
 class Example : public Base, public Mixin {
       ^
test.cpp:3:18: note:    virtual void Base::test()
     virtual void test() = 0;
                  ^

Is it possible to do this?

I know one option is to make Mixin inherit from Base, but in my case there's several derived classes and they don't share a common ancestor.

Answer 1

You cannot directly have a class override a method not of its base class. But you can sort-of of do it in a roundabout way. I'll present two such approaches - I prefer the second.

Approach 1

This is described by Daniel Paul in a post on thinkbottomup.com.au, entitled C++ Mixins - Reuse through inheritance is good... when done the right way.

In your case, this is what it would look like:

class Base {
public:
    virtual void test() = 0;
};

template <typename T>
class Mixin : public T {
public:
    virtual void test() override { /*... do stuff ... */ }
};

class UnmixedExample : public Base {
    /* definitions specific to the Example class _not_including_
       a definition of the test() method */
};

using Example = class Mixin<UnmixedExample>;

int main(int argc, char** argv) {
    Example{}.test();
    return 0;
}

Approach 2: CRTP!

CRTP is the "Curiously Recurring Template Pattern" - definitely follow that link if you haven't seen it before. With this approach, we'll be using the virtual inheritance specifier to avoid ambiguity, and unlike the previous approach - we will not be reversing the inheritance order of the Mixin and Example classes.

class Base {
public:
    virtual void test() = 0;
};

template <typename T>
class Mixin : virtual T {
public:
    virtual void test() override { /*... do stuff ... */ }
};

class Example : public virtual Base, public virtual Mixin<Base> {
    /* definitions specific to the Example class _not_including_
       a definition of the test() method */
};

int main(int argc, char** argv) {
    Example{}.test();
    return 0;
}

Note about both solutions:

• Ain't it curious how the CRTP keeps recurring all over the place? :-)
• The code I used is C++11 for pedagogical purposes, but the same would work in C++98.

Answer 2

You cannot have a class override an unrelated class's virtual function. There are different things you could do to work around this. You can make the mixin a template that derives (virtually) from the type argument and use it as class Example : public virtual Base, Mixin, or you can add code in the final class to dispatch to the mixing:

void Derived::test() { Mixin::test(); }