how to force base class constructors to be called in derived classes?

c++

Solution

Yes, the default base constructor is always called unless explicitly stated otherwise.

For example:

class A
{
public:
   A() { std::cout << "A"; }
};

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

int main()
{
   B b;
   return 0;
}

will output:

A

By "explicitly stated otherwise" I mean that you can call a different constructor from the derived class:

class A
{
public:
   A() { std::cout << "A"; }
   A(int) { std::cout << "AAA"; }
};

class B : A
{
public:
   B() : A(1) {}  //call A(int)
};

int main()
{
   B b;
   return 0;
}

will output

AAA

Important if you don't have a default constructor (you declare a non-default constructor and not a default one) or the default constructor is not visible (marked as `private`), you need to explicitly call an available constructor in the derived class.

Problem

basic c++ question i'm fairly sure. if i have a base class with a constructor that takes no parameters, and just initializes some of the protected members, does a derived class instantly call this base constructor too if it matches the parameters (wishful but unlikely thinking), and if not, is there a way to force it to automatically call said base constructor from the derived class WITHOUT having to explicitly tell it to do so in the derived class? I ask because i'm writing a wrapper of sorts and there are some protected members that i want initialized to specific values initially, and then i want to derive and manipulate this base class to my needs, but i wouldn't like an outside user to have to remember to explicitly call the base constructor or set these values within their own constructor.

Original source