How to write `is_complete` template?

After answering this question I was trying to find is_complete template in Boost library and I realized that there is no such template in Boost.TypeTraits. Why there is no such template in Boost library? How it should look like?

//! Check whether type complete
template<typename T>
struct is_complete
{   
  static const bool value = ( sizeof(T) > 0 );
};

...

// so I could use it in such a way
BOOST_STATIC_ASSERT( boost::is_complete<T>::value );

The code above is not correct, because it is illegal to apply sizeof to an incomplete type. What will be a good solution? Is it possible to apply SFINAE in this case somehow?

Well, this problem couldn't be solved in general without violating the ODR rule, but there is there a platform specific solution which works for me.

The answer given by Alexey Malistov can be used on MSVC with a minor modification:

namespace 
{
    template<class T, int discriminator>
    struct is_complete {  
      static T & getT();   
      static char (& pass(T))[2]; 
      static char pass(...);   
      static const bool value = sizeof(pass(getT()))==2;
    };
}
#define IS_COMPLETE(X) is_complete<X,__COUNTER__>::value

Unfortunately, the __COUNTER__ predefined macro is not part of the standard, so it would not work on every compiler.

It might be a bit late, but so far, no C++ 11 solution worked for both complete and abstract types.

So, here you are.

With VS2015 (v140), g++ >= 4.8.1, clang >= 3.4, this is working:

template <class T, class = void>
struct IsComplete : std::false_type
{};

template <class T>
struct IsComplete< T, decltype(void(sizeof(T))) > : std::true_type
{};

Thanks to Bat-Ulzii Luvsanbat: https://blogs.msdn.microsoft.com/vcblog/2015/12/02/partial-support-for-expression-sfinae-in-vs-2015-update-1/

With VS2013 (V120):

namespace Details
{

    template <class T>
    struct IsComplete
    {
        typedef char no;
        struct yes { char dummy[2]; };

        template <class U, class = decltype(sizeof(std::declval< U >())) >
        static yes check(U*);

        template <class U>
        static no check(...);

        static const bool value = sizeof(check< T >(nullptr)) == sizeof(yes);
    };

} // namespace Details


template <class T>
struct IsComplete : std::integral_constant< bool, Details::IsComplete< T >::value >
{};

This one is inspired from the internets and static assert that template typename T is NOT complete?

template<class T>
struct is_complete {
    static T & getT();
    static char (& pass(T))[2];
    static char pass(...);

    static const bool value = sizeof(pass(getT()))==2;
};

I'm afraid you can't implement such an is_complete type traits. The implementation given by @Alexey fails to compile on G++ 4.4.2 and G++ 4.5.0:

error: initializing argument 1 of ‘static char (& is_complete::pass(T))[2] [with T = Foo]’

On my Mac, with G++ 4.0.1 evaluating is_complete<Foo>::value where struct Foo; is incomplete yields to true which is even worse than a compiler error.

T can be both complete and incomplete in the same program, depending on the translation unit but it's always the same type. As a consequence, as commented above, is_complete<T> is always the same type as well.

So if you respect ODR it is not possible to have is_complete<T> evaluating to different values depending on where it is used; otherwise it would mean you have different definitions for is_complete<T> which ODR forbids.

EDIT: As the accepted answer, I myself hacked around a solution that uses the __COUNTER__ macro to instantiate a different is_complete<T, int> type everytime the IS_COMPLETE macro is used. However, with gcc, I couldn't get SFINAE to work in the first place.

How to write `is_complete` template?

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