virtual multiple inheritance constructor

Question

I'm working on a C++ code example that uses virtual inheritance and multiple inheritance. In my code, I've noticed that I have to call the constructor of the Base class in each derived class, even though the values passed to the base class constructor are not used. Why is this necessary, and can I avoid calling the base class constructor in every derived class and only call it in the Combined class?

#include <stdio.h>

class Base
{
    public:
        Base(int n) : n(n) {}
        int n;
};

class AddsSomething1 : virtual public Base
{
    public:
        AddsSomething1() : Base(1)
        {
            printf("AddsSomething1(): base %d\n", this->n);
        }

        int something1;
};

class AddsSomething2 : virtual public Base
{
    public:
        AddsSomething2(): Base(2)
        {
            printf("AddsSomething2(): base %d\n", this->n);
        }

        int something2;
};

class AddsSomething3 : virtual public Base
{
    public:
        AddsSomething3() : Base(3)
        {
            printf("AddsSomething3(): base %d\n", this->n);
        }

        int something3;
};

class Combined : public AddsSomething1, public AddsSomething2, public AddsSomething3
{
    public:
        Combined(int n) : Base(123)
        {
            printf("Combined(): base %d\n", this->n);
        }
};

int main()
{
    Combined(123);
}

Output:

AddsSomething1(): base 123
AddsSomething2(): base 123
AddsSomething3(): base 123
Combined(): base 123

SoronelHaetir · Accepted Answer · 2023-10-20 21:49:12Z

1

A virtual base's constructor must be invoked from every derived type because it is always and only the single most-derived type constructor that actually invokes the virtual bases constructor. Every other base in the inheritance chain learns that the virtual base has already been constructed and skips that step.

answered Oct 20, 2023 at 21:49

SoronelHaetir

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6 Comments

Captain Giraffe Over a year ago

I understand what you are trying to say but I have a hard time understanding what you mean by Every other base...

SoronelHaetir Over a year ago

Say you have struct V1 { ... }; struct B1 : virtual V1 { ... }; struct B2 : virtual V1 { ... }; struct D : B1, B2 { ... }; Without virtual there would be two V1 instances in the tree and the construction order would be V1, B1, V1, B2, D. With virtual it is V1, B1, B2, D. A type only knows that it is the most-derived by the fact that the virtual base has not yet been initialized.

mopalinski Over a year ago

Are there any exceptions to this rule? I am also playing with templates and I think I found an example where I don't have to call Base() from deriving class. Can I contact you?

SoronelHaetir Over a year ago

@PoProstuMieciek : If the virtual base has a default constructor then it does not need to be explicitly invoked. Such a constructor will, however, be invoked as normal, just by the most-derived type rather than the left-most direct base.

einpoklum Jun 4 at 22:48

What about if the ctor is invoked with different parameters?

|

阴阳战士 · Accepted Answer · 2023-10-20 21:58:08Z

0

No, you can't avoid invoking parent class constructor because they are invoked in implicity. Also, you don't need to invoke them in explicity. You know, we can't imagine the successful inheritance without invoking them.

Thank you.

answered Oct 20, 2023 at 21:58

阴阳战士

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