Thread: How constructors are called in c++ (inheritance)

struggling with constructors in c++.. why in a code below base class default constructor is called despite of constructor with parameter being already executed? In D class constructor with parameter D:(int n) : B(n) constructor is called explicitly first but after that default constructor B() is still executed before anything else and I couldnt figure out why? Thank you for any help.

Code:

class B
   { 
      public:
         B();
         B(int n);
   };

   B::B()
   {
      cout << "B::B()\n";
   }

   B::B(int n)
   {
      cout << "B::B(" << n << ")\n";
   }


   class D : public B
   { 
      public:
         D();
         D(int n);
      private:
         B b;
   };

   D::D()
   {
      cout << "D::D()\n";
   }

   D::D(int n) : B(n)
   {
      b = B(-n);
      cout << "D::D(" << n << ")\n";
   }

   int main()
   {
      D d(3);
      return 0;
   }

output is:

B::B(3)
B::B()
B::B(-3)
D:(3)

sorry about smiley above, should be :: there

>> but after that default constructor B() is still executed before anything else and I couldnt figure out why?

You have a member variable of type B inside class D. That will be constructed when a variable of type D is created. That construction happens in the initialization list. If you don't explicitly construct b inside the initialization list, the default constructor will be called before the constructor's code is executed. That means that this line:

Code:

b = B(-n);

comes after b was already default constructed. It then creates a temporary B object with -n and assigns that temporary to b.

If you created an assignment operator and added a cout output to it, you would see that that code is assigning to b, not constructing it.

Thank you for such a quick help. I cant believe I missed private b variable, now it makes perfect sense to me. Thank you again for your help.