在构造函数继承中使用默认构造函数

Question

I have a template class "Derived" which does a constructor inheritance:

template <class T>
class Derived : public T
{
    using T::T;
    Derived()
    {
         std::cout<<"in derived"; 
    }
};

My base class has a constructor that expects arguments:

class Base
{
public:
    Base(int a)
    {
        std::cout<<"in base";
    }
};

When I create an object of type Derived it seems the derived constructor is not called:

Derived<Base> derived(2);

prints "in base";

Why? Is there a way to tell it to call Derived constructor?

Answer 1

Initialization by an inherited constructor happens as follows:

[class.inhctor.init] (emphasis mine)

1 When a constructor for type B is invoked to initialize an object of a different type D (that is, when the constructor was inherited), initialization proceeds as if a defaulted default constructor were used to initialize the D object and each base class subobject from which the constructor was inherited, except that the B subobject is initialized by the invocation of the inherited constructor . The complete initialization is considered to be a single function call; in particular, the initialization of the inherited constructor's parameters is sequenced before the initialization of any part of the D object.

The key point is the defaulted word. Defaulted c'tors are generated by the compiler, for instance a copy c'tor can be defaulted. And so it doesn't use any c'tor that's defined in the derived class. A compiler generated c'tor is always going to have an empty compound statement. So one should not expect anything to be printed.