Compile-Time Polymorphism for Data Members

Question

In the following code, initialize() illustrates a method based on compile-time polymorphism. The version of initialize() compiled depends on int2type<true> and int2type<false>, only one of which will be true for a given template parameter T.

It just so happens that data member T* m_datum; will work for both int2type<true> and int2type<false>.

Now, I want to change the int2type<false> version to std::vector<T> m_datum;, so my question is, how do I modify my code so that the data member m_datum is polymorphic on int2type<>?

Note: please ignore the rationale behind the code below - instead, I would like to focus on the mechanics of achieving compile-time polymorphism for data members.

#include <type_traits>
#include <stdlib.h>

using namespace std;

template <bool n>
struct int2type
{
  enum { value = n };
};

template< typename T >
struct is_trivially_copyable
{
  static const bool value = std::is_standard_layout<T>::value;
};

template<class T>
class Foo
{
  public:
    Foo( size_t n ) : m_nr( n )
    {
      initialize( int2type<is_trivially_copyable<T>::value>() );        
    }
    ~Foo() { }

  private:
    void initialize( int2type<true> )
    {
      m_datum = (T*) calloc( sizeof(T), m_nr );
    }
    void initialize( int2type<false> )
    {
      m_datum = new T[m_nr];
    }

  private:
     size_t     m_nr;
     T*         m_datum;   // ok for int2type<true>
 //  vector<T>  m_datum;   // want to change to this for int2type<false>
};

class Bar
{
  public:
    Bar() { }
    virtual ~Bar() { }
};

int main(int argc, char** argv)
{
  Foo<int> foo_trivial(     5 );
  Foo<Bar> foo_nontrivial( 10 );

  return 0;
}

C++11 solution, based on Nawaz's recommendations

#include <type_traits>
#include <vector>
#include <stdlib.h>

using namespace std;

template< typename T >
struct is_trivially_copyable
{
    static const bool value = std::is_standard_layout<T>::value;
};

template<class T>
class Foo
{
    private:
        static const bool what = is_trivially_copyable<T>::value;
        typedef typename std::conditional<what,T*,std::vector<T>>::type type;

    public:
        Foo( size_t n ) : m_nr( n )
        {
            initialize( m_datum );      
        }
        ~Foo() { }

    private:
        void initialize( T* dummy )
        {
            m_datum = (T*) calloc( sizeof(T), m_nr );
        }
        void initialize( std::vector<T>& dummy )
        {
            m_datum.resize( m_nr );             
        }

    private:
        size_t     m_nr;
        type       m_datum;   
};

class Bar
{
    public:
        Bar()  { }
        virtual ~Bar() { }
};

int main(int argc, char** argv)
{
    Foo<int> foo_trivial(     5 );
    Foo<Bar> foo_nontrivial( 10 );

    return 0;
}

Answer 1

C++11 Solution

Use std::conditional as:

#include <type_traits>

template<class T>
class Foo
{
  //some info we can use throughout the class
  static const bool what = is_trivially_copyable<T>::value;
  typedef typename std::conditional<what, T*, std::vector<T>>::type data_type;

  //data members
  data_type m_data;  //this is what you need!
}

---

C++03 Solution

You can write a metafunction and partially specialize this as follows:

template<class T>
class Foo
{
     //primary template
     template<bool b, typename T> 
     struct get { typedef T* type; };

     //partial specialization
     template<typename T> 
     struct get<false, T> { typedef std::vector<T> type; };

     //some info we can use throughout the class
     static const bool what = is_trivially_copyable<T>::value;
     typedef typename get<what, T>::type data_type;

     //data members
     data_type m_data;  //this is what you need!
};

So when what is true, data_type will turn out to be T*, or else it will be std::vector<T>, as desired.

In either case, you don't need int2type class template. Just remove that from your code. You can write cleaner code, without it.