Question

I have class A with basic members and functions:

class A{
public:
    typedef struct{
        int aa;
        int bb;
    } stEntry;
    stEntry entry;

    void function1();
    void function2();
};

Than class B that should extend class A including structure stEntry…

class B : public A{
public:
    typedef struct :stEntry
    {
        int cc;
    } stEntry;
    stEntry entry;

    void function3();
};

and then:

int main() {
    A a;
    B b;
    a.entry.aa = 1;    
    b.entry.aa = 2;
    b.entry.cc = 3;

    cout << "class A:"<<sizeof(a)<< endl;
    cout << "class B:"<<sizeof(b)<< endl;

    return 0;
}

I get

class A:8
class B:20

So class B contains 2 instances – 8 bytes(A class member) + 12 bytes(B class member).

Is there a way how to extend structure stEntry for class B? (without have 2 instances)

Answer 1

Sort of, with virtual inheritance:

struct stEntryBase {
    int aa;
    int bb;
};

struct A : virtual stEntryBase {
    typedef stEntryBase stEntry;

    void function1();
    void function2();
};

struct stEntryDerived : virtual stEntryBase {
    int cc;
};

struct B : A, stEntryDerived {
    typedef stEntryDerived stEntry;

    void function3();
};

If you want to go another level of inheritance then B would derived virtually from stEntryDerived.

Now you have to refer to the fields as a.aa, b.cc, there is no member entry. Also, the stEntry types are no longer POD (so bb and cc may no longer be adjacent in memory). Finally, the size increase due to the virtual inheritance might actually be bigger than two ints.

What you can do, is get an stEntryDerived* or stEntryDerived& from an instance of B. Then that pointer/reference can be used to access aa, bb, and cc as the members of stEntryDerived, without the user needing to know about B. So you’ve achieved separation of interface, at some cost.