I would like to check if a certain template specialization exist or not, where the general case is not defined.
Given:
template <typename T> struct A; // general definition not defined
template <> struct A<int> {}; // specialization defined for int
I would like to define a struct like this:
template <typename T>
struct IsDefined
{
static const bool value = ???; // true if A<T> exist, false if it does not
};
Is there a way to do that (ideally without C++11)?
Thanks
Using the fact that you can't apply sizeof
to an incomplete type:
template <class T, std::size_t = sizeof(T)>
std::true_type is_complete_impl(T *);
std::false_type is_complete_impl(...);
template <class T>
using is_complete = decltype(is_complete_impl(std::declval<T*>()));
See it live on Coliru
Here is a slightly clunky, but working C++03 solution:
template <class T>
char is_complete_impl(char (*)[sizeof(T)]);
template <class>
char (&is_complete_impl(...))[2];
template <class T>
struct is_complete {
enum { value = sizeof(is_complete_impl<T>(0)) == sizeof(char) };
};
See it live on Coliru
This is an alternative implementation always using the same trick @Quentin used
C++11 version
template<class First, std::size_t>
using first_t = First;
template<class T>
struct is_complete_type: std::false_type {};
template<class T>
struct is_complete_type<first_t<T, sizeof(T)>> : std::true_type {};
Example on wandbox
Tentative C++03 version which does not work
template<typename First, std::size_t>
struct first { typedef First type; };
template<typename T>
struct is_complete_type { static const bool value = false; };
template<typename T>
struct is_complete_type< typename first<T, sizeof(T)>::type > { static const bool value = true; };
The error in this case is
prog.cc:11:8: error: template parameters not deducible in partial specialization:
struct is_complete_type< typename first::type > { static const bool value = true; };
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
prog.cc:11:8: note: 'T'