Multiple variable template specializations with std::enable_if - c++

Question

I'm trying to concisely define a variable template with these effective values:
// (template<typename T> constexpr T EXP = std::numeric_limits<T>::max_exponent / 2;)
// float and double scalar definitions:
const double huge = std::scalbn(1, EXP<double>);
const float huge = std::scalbn(1, EXP<float>);
// SIMD vector definitions:
const Vec8f huge = Vec8f(huge<float>); // vector of 8 floats
const Vec8d huge = Vec8d(huge<double>); // vector of 8 doubles
const Vec4f huge = Vec4f(huge<float>); // vector of 4 floats
// Integral types should fail to compile
The VecXX vector definitions (SIMD vectors) need to use the corresponding scalar type as shown (e.g. huge<float> for vector of floats). This is available as VecXX::value_type or through a type traits-style template class (VectorTraits<VecXX>::value_type).
Ideally I think I'd have something like:
// Primary. What should go here? I want all other types to not compile
template<typename T, typename Enabler = void>
const T huge = T{ 0 };
// Scalar specialization for floating point types:
template<typename T>
const T huge<T> = std::enable_if_t<std::is_floating_point<T>::value, T>(std::scalbn(1, EXP<T>));
// Vector specialization, uses above declaration for corresponding FP type
template<typename T>
const T huge<T> = std::enable_if_t<VectorTraits<T>::is_vector, T>(huge<VectorTraits<T>::scalar_type>);
but I can't quite figure out a working version (above fails with "redefinition of const T huge<T>"). What's the best way to do this?

Answer 1

Not exactly what you asked but I hope the following example can show you how to use SFINAE to specialize a template variable
template <typename T, typename = void>
constexpr T huge = T{0};
template <typename T>
constexpr T huge<T, std::enable_if_t<std::is_floating_point<T>{}>> = T{1};
template <typename T>
constexpr T huge<std::vector<T>> = T{2};
You can check it with
std::cout << huge<int> << std::endl;
std::cout << huge<long> << std::endl;
std::cout << huge<float> << std::endl;
std::cout << huge<double> << std::endl;
std::cout << huge<long double> << std::endl;
std::cout << huge<std::vector<int>> << std::endl;

Answer 2

Leaving #max66's answer as the accepted one for the credit, but here's the specific solution I wound up with:
struct _VectorTraits { static constexpr bool is_vector = true; };
template<class T> struct VectorTraits : _VectorTraits { static constexpr bool is_vector = false; };
template<> struct VectorTraits<Vec4f> : _VectorTraits { typedef float value_type; };
template<> struct VectorTraits<Vec8f> : _VectorTraits { typedef float value_type; };
template<> struct VectorTraits<Vec4d> : _VectorTraits { typedef double value_type; };
template<typename T> using EnableIfFP = std::enable_if_t<std::is_floating_point<T>::value>;
template<typename T> using EnableIfVec = std::enable_if_t<VectorTraits<T>::is_vector>;
template<typename T> constexpr T EXP = std::numeric_limits<T>::max_exponent / 2;
// Actual variable template, finally:
template<typename T, typename Enabler = void> const T huge = T{ 0 };
template<typename T> const T huge<T, EnableIfFP<T> > = std::scalbn(1, EXP<T>);
template<typename T> const T huge<T, EnableIfVec<T> > = T{ huge<typename VectorTraits<T>::value_type> };
This could still be improved I think:
It's verbose. T appears 4 times in each specialization's left-hand side.
Integral types (e.g. huge<uint32_t>) still compile with a nonsense 0 value. I'd rather them not compile.