Inherit interfaces which share a method name

This problem doesn’t come up very often. The solution I’m familiar with was designed by Doug McIlroy and appears in Bjarne Stroustrup’s books (presented in both Design & Evolution of C++ section 12.8 and The C++ Programming Language section 25.6). According to the discussion in Design & Evolution, there was a proposal to handle this specific case elegantly, but it was rejected because “such name clashes were unlikely to become common enough to warrant a separate language feature,” and “not likely to become everyday work for novices.”

Not only do you need to call Name() through pointers to base classes, you need a way to say which Name() you want when operating on the derived class. The solution adds some indirection:

class Interface1{
public:
    virtual void Name() = 0;
};

class Interface2{
public:
    virtual void Name() = 0;
};

class Interface1_helper : public Interface1{
public:
    virtual void I1_Name() = 0;
    void Name() override
    {
        I1_Name();
    }
};

class Interface2_helper : public Interface2{
public:
    virtual void I2_Name() = 0;
    void Name() override
    {
        I2_Name();
    }
};

class RealClass: public Interface1_helper, public Interface2_helper{
public:
    void I1_Name() override
    {
        printf("Interface1 OK?\n");
    }
    void I2_Name() override
    {
        printf("Interface2 OK?\n");
    }
};

int main()
{
    RealClass rc;
    Interface1* i1 = &rc;
    Interface2* i2 = &rc;
    i1->Name();
    i2->Name();
    rc.I1_Name();
    rc.I2_Name();
}

Not pretty, but the decision was it’s not needed often.