Is it possible to inherit the implementation of an interface

Philinator Published at Dev

Philinator

is it possible to inherit the implementation of an interface without implementing it for each implementation?

In the image above IBase and Ichild are classes having only pure virtual methods and Base implements the methods of IBase and Child implements the methods of IChild. Is it now somehow possible for Child to inherit the implementations of the methods of IBase via Base, so that Child does not have to implement these methods itself?

I tried to implement it like this:

struct IBase {
    virtual void do_something_Base() = 0;
};

struct IChild : public virtual IBase {
    virtual void do_something_Child() = 0;
};


struct Base : public virtual IBase {
    virtual void do_something_Base() { //implementation IBase method
        //do sth
    }
};

struct Child : public Base , public IChild {
    virtual void do_something_Child() { //implementation IChild method
        //do sth
    }
};

int main() {

    IBase* B = new Child;
    B->do_something_Base();

    delete B;

    return 0;
}

The problem with this code is that the copiler gives a warning:

warning C4250: 'Child': inherits 'Base::Base::do_something_Base' via dominance

Futhermore the program crashes at the line delete B. (If I use IChild* B instead of IBase* B the program does not crash. why is that?)

Thanks in advance

Angew is no longer proud of SO

I've been using a similar setup in a project, and it works for me. It should be enough to do the following:

Give IBase a virtual destructor. A virtual destructor is necessary for being able to delete through pointers to a base class.
To be on the safe side, derive Child from IChild virtually as well. Since the I-classes have no data, there's no point in using them as non-virtual bases.
Disable the warning, as it's basically telling that the compiler is doing exactly what you want it to do (that the implementation is inherited from the correct base class).

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edited at2021-02-21

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