CODEWITHSUNDEEP
c++templatesinterface
Jesús Enrique
Jesús Enrique

Reputation: 71

how to use an interface to force a generic datatype from a template in C++?

Good morning, I've already searched for a few things but can't get the answer. I'm creating a generic class, and the idea is that generic data tye (typename T) should come from an interface:

this is the interface:

template<typename T> class iDataType{

public:

   virtual bool writeOnFile(std::fstream& theFile,T& data) = 0;
   virtual T readOnFile(std::fstream& theFile) = 0;

};

And this is what I'm 'trying' to do:

template <typename T : public iDataType> class Database{};

thank you for your time.

Upvotes: 4

Views: 277

Answers (2)

Jeff Garrett
Jeff Garrett

Reputation: 7528

The requirement that T derive from iDataType<T> in the limited context of the question does not seem to add value. But, to answer the question as I interpret it, the property you want to be true is:

std::is_convertible_v<T*, iDataType<T>*>

std::is_base_of is misleading in that it will answer whether T has a base of iDataType<T>, but the language requires a public unambiguous base for a T& to be used as an iDataType<T>&. So, in C++, when you want to know if something has a base class as an interface, std::is_base_of is wrong.

Upvotes: 3

Alex Guteniev
Alex Guteniev

Reputation: 13749

You can assert on compile-time condition with static_assert, and there is std::is_base_of.

So the solution is:

template <typename T> class Database{
   static_assert(std::is_base_of<iDataType<T>, T>::value, "should be derived from iDataType");
};

Note that the solution is for C++11, for later C++ Standard you can use std::is_base_of_v and terse static_assert (without string literal parameter).

Upvotes: 3

Related Questions