Why are member classes initialized first in C++?

Question

I have tried the ff. code:

#include <iostream>
struct A
{
    A() { std::cout << "1"; }
    A(const A&) { std::cout << "2"; }
    A(A&&) { std::cout << "3"; }
};

struct B
{
    B() { std::cout << "4"; }
    B(const B& b) : a(b.a) { std::cout << "5"; }
    B(B&& b) : a(b.a) { std::cout << "6"; }
    A a;
};

int main()
{
    B b1;
    std::cout << "END OF b1" << std::endl;
    B b2 = std::move(b1);
}

And the output is:

14END OF b1
26

I was just wondering why that is the behavior, how come the constructor of B is called second? I've also tried to use class instead of struct and it's the same behavior.

Answer 1

This is expected behavior. The initialization order is specified as

1) If the constructor is for the most-derived class, virtual base classes are initialized in the order in which they appear in depth-first left-to-right traversal of the base class declarations (left-to-right refers to the appearance in base-specifier lists)

2) Then, direct base classes are initialized in left-to-right order as they appear in this class's base-specifier list

3) Then, non-static data members are initialized in order of declaration in the class definition.

4) Finally, the body of the constructor is executed

Then the data member a is always initialized (step #3) before the execution of the constructor of B (step #4).