Use enable_if with is_integral to make distribution traits - c++

Question

I want to make a traits for std::uniform_*_distribution according to type given. E.g.:
distribution_traits<float>::type int_dist;
I tried following ways, but none of them compiles, and I don't know why.
Implementation 1
Use std::enable_if with typedefs:
template <typename T>
struct distribution_traits {
using type = typename std::enable_if<std::is_integral<T>::value, std::uniform_int_distribution<T>>::type;
using type = typename std::enable_if<std::is_floating_point<T>::value, std::uniform_real_distribution<T>>::type;
};
Clang 3.4 complains:
dist_traits.cpp:7:9: error: redefinition of 'type'
using type = typename std::enable_if<std::is_floating_point<T>::value, std::uniform_real_distribution<T>>::type;
^
dist_traits.cpp:6:9: note: previous definition is here
using type = typename std::enable_if<std::is_integral<T>::value, std::uniform_int_distribution<T>>::type;
^
dist_traits.cpp:6:40: error: no type named 'type' in 'std::enable_if<false, std::uniform_int_distribution<float> >'; 'enable_if' cannot be used to
disable this declaration
using type = typename std::enable_if<std::is_integral<T>::value, std::uniform_int_distribution<T>>::type;
^~~~~~~~~~~~~~~~~~~~~~~~~~
dist_traits.cpp:28:3: note: in instantiation of template class 'distribution_traits<float>' requested here
distribution_traits<float>::type int_dist;
^
2 errors generated.
Implementation 2
Use enable_if as class template parameter:
template <typename T, typename distribution_t = void>
struct distribution_traits;
template <typename T>
struct distribution_traits<
T, typename std::enable_if<std::is_integral<T>::value,
std::uniform_int_distribution<T> >::type > {
using type = std::uniform_int_distribution<T>;
};
template <typename T>
struct distribution_traits<
T, typename std::enable_if<std::is_floating_point<T>::value,
std::uniform_real_distribution<T> >::type > {
using type = std::uniform_real_distribution<T>;
};
And Clang complains
dist_traits.cpp:28:3: error: implicit instantiation of undefined template 'distribution_traits<float, void>'
distribution_traits<float>::type int_dist;
^
Either way cannot be compiled by MSVC++ 12.0, and the error messages are similar.
Could anyone please explain what's wrong I'm doing with SFINAE? Thanks!
For those who are curious about solution, here is the one that compiles:
template <typename T>
auto dist() -> typename std::enable_if<std::is_integral<T>::value, std::uniform_int_distribution<T>>::type;
template <typename T>
auto dist() -> typename std::enable_if<std::is_floating_point<T>::value, std::uniform_real_distribution<T>>::type;
template <typename T>
struct distribution_traits {
using type = decltype(dist<T>());
};
BTW, if put dist function into distribution_traits, the compilation will fail with error: function only differs in return type cannot be overloaded. :(

Answer 1

SFINAE can be used to discard overloads of function templates and class template specializations during substitution of template arguments.
It cannot be used with type/template aliases like you're trying to do.
About your working code - putting dist inside the class doesn't work because you attempt to call dist inside decltype without an object. Make dist static and it'll work:
template <typename T>
struct distribution_traits {
template <typename U>
static auto dist() -> typename std::enable_if<std::is_integral<U>::value, std::uniform_int_distribution<U>>::type;
template <typename U>
static auto dist() -> typename std::enable_if<std::is_floating_point<U>::value, std::uniform_real_distribution<U>>::type;
using type = decltype(dist<T>());
};
For implementation 2 to work, you need to omit the second argument of enable_if:
template
struct distribution_traits;
template <typename T>
struct distribution_traits<
T, typename std::enable_if<std::is_integral<T>::value>::type> {
using type = std::uniform_int_distribution<T>;
};
otherwise the specialization you define is distribution_traits<T, uniform_int_distribution<T>> and that doesn't match an instantiation like distribution_traits<float> because the second parameter is defaulted to void.