How to decide if a template specialization exist

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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

template-specialization sfinae templates c++
2 Answers

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'

user_6119010
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