member variable initialization

Hello everyone,


I am wondering in the following code, member variable a in class B is not put in the initialization list or constructor of B directly, but it is initialized. How and when member variable a of class B is created and initialized? Is constructor of B invokes constructor of a?

Output is,

In constructor A
In constructor B

Code:

---

#include <iostream>

using namespace std;

class A
{
public:

        A()
        {
                cout << "In constructor A" << endl;
        }
};

class B
{
public:

        A a;

        B()
        {
                cout << "In constructor B" << endl;
        }
};

int main()
{
        B b;

        return 0;
}

---

thanks in advance,
George

Yes, all member variables that are of a class type are implicitly called [if not explicitly called] when the instance is constructed. This is part of the basics of object oriented programming - since you can't actually know (necessarily) the base class of your current object if it's inherited from another class, so you you wouldn't know the chain of constructors that you would need to call - in particular since the first call has to be to the most basic constructor. Say for example you have a "ifstream" in your class, you wouldn't know what the base-class(es) is(are) for that class, right? But the compiler can figure it out and call the constructor.

--
Mats

Code:

---

        B()
        {
                cout << "In constructor B" << endl;
        }

---

is equivalent to:

Code:

---

        B() : a()
        {
                cout << "In constructor B" << endl;
        }

---

The compiler automatically calls the default constructor of a member object if you don't explicitly initialize it in the initializer list.

Thanks Mats,


My question is answered.

Quote:

---

Originally Posted by matsp

Yes, all member variables that are of a class type are implicitly called [if not explicitly called] when the instance is constructed. This is part of the basics of object oriented programming - since you can't actually know (necessarily) the base class of your current object if it's inherited from another class, so you you wouldn't know the chain of constructors that you would need to call - in particular since the first call has to be to the most basic constructor. Say for example you have a "ifstream" in your class, you wouldn't know what the base-class(es) is(are) for that class, right? But the compiler can figure it out and call the constructor.

--
Mats

---

regards,
George

Thanks Daved,


My question is answered.

Quote:

---

Originally Posted by Daved

Code:

---

        B()
        {
                cout << "In constructor B" << endl;
        }

---

is equivalent to:

Code:

---

        B() : a()
        {
                cout << "In constructor B" << endl;
        }

---

The compiler automatically calls the default constructor of a member object if you don't explicitly initialize it in the initializer list.

---

regards,
George

Also, in anticipation of your next question, member variables are always initialized in the order that they are declared in the class declaration, not in the order they are specified in member initialization lists. For example:

Code:

---

class Test
{
public:
    Test() : m_Str(), m_List(), m_Vec() {}

private:
    std::list<double>  m_List;
    std::string        m_Str;
    std::vector<int>  m_Str;
};

---

In this example, m_List will be initialized before m_Str, even though the constructor makes it look like m_Str is being initialized first.

And a proper compiler will warn you about it.