Is there a more convenient solution for enable_if error in CLang (error 11723)?

Question

Ideally, we can do something like this with enable_if:

#include <type_traits>

namespace detail {
    enum class enabler_t { DUMMY };
}

template<bool B>
using enable_if_u = typename std::enable_if<B, detail::enabler_t>::type;

template<bool B>
using disable_if_u = typename std::enable_if<!B, detail::enabler_t>::type;

template<typename T, enable_if_u<std::is_same<T, int>::value>...>
int a(){return 0;}

template<typename T, disable_if_u<std::is_same<T, int>::value>...>
double a(){return 0.0;}

int main() {
    auto x = a<int>();
}

Which, imho, is the best way to use it.

However, this does not work with Clang due to error 11723 .

The workaround I'm using is using the constexpr variable:

namespace detail {
    enum class enabler_t { DUMMY };
    constexpr const enabler_t dummy = enabler_t::DUMMY;
}

//...

template<typename T, enable_if_u<std::is_same<T, int>::value> = detail::dummy>
int a(){return 0;}

template<typename T, disable_if_u<std::is_same<T, int>::value> = detail::dummy>
double a(){return 0.0;}

//...

I am not a big fan of this workaround. I was wondering if there is another way to work around code in Clang? I am looking for a clean solution in C ++, without using a preprocessor.

The example itself is here to illustrate the problem, it is not useful at all, and there are tons of ways to do it better. I just want to find a more suitable solution for the enable_if part.

Answer 1

:

#define ENABLE_IF(...) typename std::enable_if<(__VA_ARGS__), detail::enabler_t>::type=detail::enabler_t::DUMMY

, ++, . , , ZLang, , :

#define BW_ENABLE_IF(...) typename std::enable_if<(__VA_ARGS__), detail::enabler_t>::type=detail::enabler_t::DUMMY
#define ZLANG_bw_enable_if (BW_ENABLE_IF)

, -DZLANG_NS=bw:

template<typename T, $(enable_if std::is_same<T, int>())>
int a(){return 0;}