Question

#include <memory>

class Base
{
    std::shared_ptr<Base> create() const; // Returns a default constructed object
}

Suppose, that all the members derived to whatever degree from Base are copy constructible and default constructible. I want the

std::shared_ptr<Base> create() const;

Method to create object of the appropriate dynamic type, but I do not want to use boilerplate code.

Is it possible to make

std::shared_ptr<Base> create() const;

Statically bound, but inside find somehow the correct type and create the object using Default constructor? Possibly using C++11.

Answer 1

The create() functions should probably be static, as you don’t have an instance yet. But without parameters you cannot do what you want… unless you use templates, of course:

class Base
{
public:
    template<typename T>
    static std::shared_ptr<Base> create() const
    {
        return std::shared<Base>(new T);
    }
};

Then use it this way:

std::shared_ptr<Base> ptr(Base::create<Foo>());

Or, if you prefer:

std::shared_ptr<Base> ptr(Foo::create<Foo>());