How to access base class constructor when using CRTP

Question

I need to insert clone and create member functions to my class hieararchy

class Base
{
protected:
    const int x_;
public:
    Base() : x_(0) {}
    Base(int x) : x_(x) {}
};

I thought that CRTP could be the way how to save some typing and avoid errors.

template <typename Derived>
class CRTP_Iface : public Base
{
public:
    virtual Base *create() const { return new Derived(); }
    virtual Base *clone() const { return new Derived(static_cast<Derived const&>(*this)); }
};

Unfortunately I'm not able to access base class constructor to initialize const members.

class D1 : public CRTP_Iface<D1>
{
public:
    D1() : Base() {}
    D1(int x) : Base(x) {}
};

class D2 : public CRTP_Iface<D2>
{
public:
    D2() : x_(0) {}
    D2(int x) : x_(x) {}
};

int main()
{
    D1 a;
    D2 b;

    return 0;
}

Is there any simple way how to solve this?

Answer 1

Simply add all needed constructors to CRTP_Iface.

public:
  CRTP_Iface() : Base() {}
  CRTP_Iface( int x ) : Base(x) {}

If using C++11 this gets even easier:

public:
  using Base::Base;

Then you have:

class D1 : public CRTP_Iface<D1>
{
public:
    D1() : CRTP_Iface() {}
    D1(int x) : CRTP_Iface(x) {}
};

... which can be nicer written in C++11:

class D1 : public CRTP_Iface<D1>
{
public:
  using CRTP_Iface<D1>::CRTP_Iface;
};

(not sure if is needed on either left hand or right hand of ::, AFAIR some complilers like it more strict)

Answer 2

You can inherit the constructors of class Base in template CRTP_Iface<> and then call its constructor in the derived classes:

template <typename Derived>
class CRTP_Iface : public Base
{
protected:
    using Base::Base;

public:
    virtual Base *create() const { return new Derived(); }
    virtual Base *clone() const { return new Derived(static_cast<Derived const&>(*this)); }
};

class D1 : public CRTP_Iface<D1>
{
public:
    D1() : CRTP_Iface() {}
    D1(int x) : CRTP_Iface(x) {}
};

live example