Why is constructor of a grand parent deleted in this case?

Question

The compiler is complaining the constructor of D is deleted because of ill forming why ?

#include<iostream>
using namespace std;

class A
{
   int x;
   public:
   A(int i) { x = i; }
   void print() { cout << x; }
};

class B: virtual public A
{
   public:
      B():A(10) { }
};

class C: virtual public A 
{
   public:
      C():A(10) { }
};

class D: public B, public C {
};

int main()
{
   D d;
   d.print();
   return 0;
}

Output

main.cpp:37:4: error: use of deleted function 'D::D()' D d; ^ main.cpp:32:7: note: 'D::D()' is implicitly deleted because the default definition would be ill-formed: class D: public B, public C { ^

Because there's no default constructor available for `D` to initialize those base classes. — πάντα ῥεῖ, Jul 20 '18 at 17:49
Because of virtual inheritance `D` must call ctor of `A` directly and there is no default one there — Slava, Jul 20 '18 at 17:50
Not direct dupe but could be helpfull https://stackoverflow.com/questions/17184514/final-class-implementation-in-c — Slava, Jul 20 '18 at 18:00

R Sahu · Accepted Answer · 2018-07-20T18:09:04.760

6

Due to the rules for initialization of virtual base classes,

class D: public B, public C {
};

is equivalent to:

class D: public B, public C {
    public:
       D() : A(), B(), C() {}
};

That's why you cannot create in instance of D.

Solution 1

Change A so it has a default constructor.

class A
{
     int x;
   public:
     A(int i = 0) { x = i; }
     void print() { cout << x; }
};

Solution 2

Change D to:

class D: public B, public C {
    public:
       D() : A(0), B(), C() {}
};

or a simpler version,

class D: public B, public C {
    public:
       D() : A(0) {}
};

edited Jul 20 '18 at 18:09

answered Jul 20 '18 at 17:51

R Sahu

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2

I do not think it is necessary to call default ctors of `B` and `C` explicitly – Slava Jul 20 '18 at 17:54

Hatted Rooster · Answer 2 · 2018-07-20T17:53:53.493

2

That's because D inherits from A indirectly using virtual. A doesn't have a parameterless constructor so a compiler-generated constructor for D can't be made.

edited Jul 20 '18 at 17:53

answered Jul 20 '18 at 17:51

Hatted Rooster

35,759
6
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score 0 · Answer 3 · answered Jul 20 '18 at 18:28

Note: this is mostly just adding a reference to the standard, in case anybody might care (but as usual for him, @R. Sahu's answer is quite accurate).

The standard specifies ([class.base.init]/13) that:

In a non-delegating constructor, initialization proceeds in the following order:
(13.1) — First, and only for the constructor of the most derived class (6.6.2), virtual base classes are initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where “left-to-right” is the order of appearance of the base classes in the derived class base-specifier-list.
(13.2) — Then, direct base classes are initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).

So, since A is a virtual base class, it's initialized directly by the most derived class (D). Only afterward, the direct base classes are initialized--but for anything to compile, the most derived class must be able to initialize the virtual base class(es).

There is one point some might find interesting in a case like this. Let's modify your class structure just a tiny bit, so we to the necessary initialization, and (importantly) initialize with a unique value in each constructor:

#include <iostream>

class A {
    int i;
 public:
    A(int i) : i(i) {}

    void show() { std::cout << "value: " << i << "\n"; }
};

class B : virtual public A{
public:
    B() : A(10) {}
};

class C : virtual public A { 
public:
    C() : A(20) {}
};

class D : public B, public C {
public:
    D() : A(0) {}
};

int main() {
    D d;
    d.show();
}

In this case, what exactly happens? We have three different constructors each "thinking" it's going to initialize the A object with a different value? Which one "wins"?

The answer is that the one in the most-derived constructor (D::D) is the one that' used to initialize the virtual base class object, so that's the one that "wins". When we run the code above, it should print 0.

Why is constructor of a grand parent deleted in this case?

3 Answers3