Question

class A : public B
{
  ...
}

// case I : explicitly call the base class default constructor
A::A() : B()
{
  ...
}

// case II : don't call the base class default constructor
A::A() // : B()
{
  ...
}

Is the case II equal to case I?

To me, I assume that the default constructor of base class B is NOT called in case II. However, despite still holding this assumption, I have run a test which proves otherwise:

class B
{
public:
    B()
    {
        cout << "B constructor" << endl;
    }
};

class A : public B
{
public:
    A()
    {
        cout << "A constructor" << endl;
    }
};

int _tmain(int argc, _TCHAR* argv[])
{
    A a;
    return 0;
}

// output from VS2008

B constructor
A constructor
Press any key to continue . . .

Answer 1

The base class constructor is called in both cases.

Here is a link to an article with more info.