C++ CRTP constructor of derived class

Question

I have some questions about the CRTP. Let's say I have the following code

#include <iostream>

// interface

template<class Imp>
class Interface{
  public:
    inline void print(void){
      std::cout<<"value: ";
      return asImp().print();
    }
  private:
    typedef Imp Implementation;
    inline Implementation& asImp(void){return static_cast<Implementation&>(*this);}
};

// add
class Add:public Interface<Add>{
  public:
    inline void print(void){std::cout<<value<<std::endl;++value;}
  private:
    int value;
};

// main
int main(void){
  Interface<Add> foo;
  foo.print();
  foo.print();
}

The output is

value: 0
value: 1

Therefore the variable value seems to be constructed as 0 by the default constructor. But I don't understand where and when this constructor is called since no object of the derived class is created.

Moreover, let's suppose that I want create value with a different starting value, how can I achieve that using this design pattern? Obviously I could create an init() method in the derived class which is called in the constructor of the base class but it wouldn't work for a type which hasn't a default constructor.

Finally, is it possible to forward some arguments pass to the constructor of the base class to the constructor of the derived class?

Answer 1

You're NOT creating a derived object in fact.

You're creating a base class object and casting it to a reference of the derived type, but the underlying object is still a base class one and thus it's wrong.

static_cast always succeeds, but will raise undefined-behavior if you don't cast to the right type while dynamic_cast returns a NULL pointer if you casted it wrong.

-> Try it live here, you might or might not get garbage values: http://ideone.com/dP3jjU

The other initialization question should be straightforward once you've addressed the above.

More on this: Should static_cast<Derived *>(Base pointer) give compile time error?

Answer 2

In your source code, you yourself name the base class as Interface. In the object oriented sense, you do not create instances of an Interface, but instances of classes that derive from the Interface. Your code in main erroneously instantiates an Interface rather than the class that derived from it. You can fix this by forcing Interface with properties that make it impossible to instantiate. E.g.:

template<class Imp>
class Interface {
    //...
protected:
    Interface () {} // not accessible except by derived
};

Unlike regular polymorphism, you are not really expected to pass Interface objects around. The Interface provides enforcement of a particular interface by providing the expected methods, and those that derive from the Interface must adopt the expectations of that interface. Your example is a little contrived because the Interface is actually just dispatching to the same named method in the derived. But, a better example is one where the Interface provides an implementation using properties it expects the derived type to provide.

In the example below, we see that Worker inherits the interface of WorkerInterface. But, because of the expectations of WorkerInterface, it is required to implement perform() and wait(). While a pure interface enforces this requirement with pure virtual methods, CRTP enforces this with template expansion.

template <typename JOB>
class WorkerInterface {
public:
    void work () { while (job().wait()) job().perform(); }
private:
    JOB & job () { return *static_cast<JOB *>(this); }
protected:
    WorkerInterface () {}
};

class Worker : public WorkerInterface<Worker>
{
    friend class WorkerInterface<Worker>;
    int state_;
    void perform () { std::cout << "Worker: " << __func__ << '\n'; }
    bool wait () { return state_--; }
public:
    Worker () : state_(1) {}
};

int main ()
{
    Worker w;
    w.work();
}

Now, any class that derives from WorkerInterface will be provided a work() method that will do the "right" thing as long as the derived class provides suitable implementations of wait() and perform().