How to detect if a type is shared_ptr at compile time

Question

I want to get a templatized way of finding if a type is a shared_ptr and based on that I want to have a new specialization of a function.

Example main function is,

template <class T> inline
void CEREAL_LOAD_FUNCTION_NAME( RelaxedJSONInputArchive & ar,    NameValuePair<T> & t )
{
    std::cout << " CEREAL_LOAD_FUNCTION_NAME NameValuePair 1 " << std::endl;
     ar.setNextName( t.name );
     ar( t.value );
}

If t.value is shared_ptr then I want to have a different function specialization. I have tried below,

template <class T> inline
typename std::enable_if<is_pointer<T>::value, void>::type
CEREAL_LOAD_FUNCTION_NAME( RelaxedJSONInputArchive & ar, NameValuePair<T> & t )
 {
    std::cout << " CEREAL_LOAD_FUNCTION_NAME NameValuePair 2 " << std::endl;
   ar.setNextName( t.name );
   ar( t.value );
  }

But it does not seem to work. These are part of c++11 cereal library. Which I am trying to customize.

Answer 1

the following may help:

template<typename T> struct is_shared_ptr : std::false_type {};
template<typename T> struct is_shared_ptr<std::shared_ptr<T>> : std::true_type {};

then you can do the following to get the correct function:

template <class T> 
typename std::enable_if<is_shared_ptr<decltype(std::declval<T>().value)>::value, void>::type
func( T t )
{
    std::cout << "shared ptr" << std::endl;
}

template <class T> 
typename std::enable_if<!is_shared_ptr<decltype(std::declval<T>().value)>::value, void>::type
func( T t )
{
    std::cout << "non shared" << std::endl;
}

live demo