Question

I’ve just tried this piece of code:

struct FaceOfPast
{
    virtual void Smile() = 0;
};

struct FaceOfFuture
{
    virtual void Smile() = 0;
};

struct Janus : public FaceOfPast, public FaceOfFuture
{
    virtual void Smile() {printf(":) ");}
};

…

void main()
{
    Janus* j = new Janus();
    FaceOfFuture* future = j;
    FaceOfPast* past = j;

    future->Smile();
    past->Smile();

    delete j;
}

It works as intended (outputs two smiley faces), but I don’t think it should even compile, redeclaration of Smile() in Janus being ambiguous.

How (and why) does it work?

Answer 1

There’s no ambiguity because you call Smile() on pointers to FaceOfFuture and FaceOfPast that only declare one method Smile().

Because calling the method on a base class pointer can’t result in an ambiguity, let’s treat the situations when you call the method directly on the child class pointer:

Janus* j = new Janus();
j->Smile();

The derived class, besides overriding, also hides the base classes’ declaration of Smile(). You’d have an ambiguity only if you wouldn’t be overriding the method in your derived class:

The following compiles:

struct FaceOfPast
{
    virtual void Smile() {printf(":) ");}
};
struct FaceOfFuture
{
    virtual void Smile() {printf(":) ");}
};
struct Janus : public FaceOfPast, public FaceOfFuture
{
   virtual void Smile() {printf(":) ");}
};
int main()
{
   Janus* j = new Janus();
   j->Smile();
}

Although you call Smile on a Janus, the base class declarations are hidden.

The following doesn’t:

struct FaceOfPast
{
    virtual void Smile() {printf(":) ");}
};

struct FaceOfFuture
{
    virtual void Smile() {printf(":) ");}
};

struct Janus : public FaceOfPast, public FaceOfFuture
{
};

int main()
{
   Janus* j = new Janus();
   j->Smile();
}

Because of the ambiguity.