Is Compile-time Polymorphism in C++ Templates Possible?

Question

Consider the following code snippet:

#include <iostream>
#include <typeinfo>


template<typename T>
class TypePrinter{
    public:
    void print_type() {
        std::cout << typeid(T).name() << std::endl;
    }
}; // end class TypePrinter


class Base{
    public:
    TypePrinter<Base> createTypePrinter(){ // TODO replace "Base" with the class of the instance.
        return TypePrinter<Base>();
    }
}; // end Base

class Derived: public Base{
};

int main() {
    // Create a derived instance and try to create a TypePrinter for it
    Derived d;
    auto tp = d.createTypePrinter();
    tp.print_type(); // prints "4Base".
    return 0;
} // end main

In the example above, I have

• a class with a typename template parameter (TypePrinter) and
• another class that instantiate it, using its type as a parameter (Base)

How can the code above be rewritten so that the inherited derived_instance.createTypePrinter() method can create a TypePrinter<Derived> rather than a TypePrinter<Base>?

In other words, what modifications could allow d.createTypePrinter().print_type() to print "Derived"?

Answer 1

As noted in the comments, this can be implemented with the Curiously Recurring Template Pattern.

#include <iostream>
#include <typeinfo>


template<typename T>
class TypePrinter{
    public:
    void print_type() {
        std::cout << typeid(T).name() << std::endl;
    }
}; // end class TypePrinter

template<typename T> // Base is now a template class. 
                     // Subclasses set T to their type
class Base{
    public:
    TypePrinter<T> createTypePrinter(){
        return TypePrinter<T>();
    }
}; // end Base

class Derived: public Base<Derived>{
};

int main() {
    // Create a derived instance and try to create a TypePrinter for it
    Derived d;
    auto tp = d.createTypePrinter();
    tp.print_type(); // Now this prints "7Derived"
    return 0;
} // end main