Avoid calling base constructor 2 times

Question

Say I have several classes that inherit like so.

#include <iostream>

struct A {
    A() {std::cout << "a";}
};

struct B : A {};
struct C : A {};

struct D : B, C {};

int main {
    D d;
}

On executing the program, as expected, I see two A objects were constructed, one for the B and one for the C object that are constructed when creating a D object. Whoever, how can I not create two A objects? I want the same A object to be used to create B and C objects. Is this possible?

Possible duplicate of [C++, are multiple-inherited constructors called multiple times?](http://stackoverflow.com/questions/7405839/c-are-multiple-inherited-constructors-called-multiple-times) — Danh, Nov 03 '16 at 10:39

score 7 · Accepted Answer · edited May 23 '17 at 12:06

7

If B and C both use virtual inheritance for A, then there will only be a single base class object for each D object:

struct B : virtual A {};
struct C : virtual A {};

//...
D d; //prints "a" rather than "aa"

Live demo

edited May 23 '17 at 12:06

Community

1
1

answered Nov 03 '16 at 10:40

TartanLlama

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score 2 · Answer 2 · answered Nov 03 '16 at 10:41

2

If you're looking for a non-polymorphic solution, you're out of luck. Otherwise, change to the following:

struct B : virtual A {};
struct C : virtual A {};

answered Nov 03 '16 at 10:41

DeiDei

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score 1 · Answer 3 · answered Nov 03 '16 at 17:24

As written, every D object has two sub-objects of type A, one inherited from class B and one inherited from class C. So the constructor of A must run twice, once for each sub-object.

If the original design is wrong, and there should only be one A sub-object, despite the use of two bases B and C, the change is to make A a virtual base of B and a virtual base of C. That way there will only be one A sub-object in a D object, and the constructor of A will only run once when a D object is constructed.

Avoid calling base constructor 2 times

3 Answers3