How to port C++ std::is_reference<T> to D ?

[wjoe]

On Wednesday, 6 May 2020 at 09:19:10 UTC, drug wrote:
> 06.05.2020 12:07, wjoe пишет:
>> Hello,
>> 
>> I'm choking on a piece of C++ I have no idea about how to 
>> translate to D.
>> 
>>    template <typename T,
>>          typename std::enable_if< std::is_const<T>::value == 
>> true, void>::type* = nullptr>
>>      constexpr const char *modifier() const {
>>          return "[in] ";
>>      }
>> 
>>    template <typename T,
>>          typename std::enable_if< std::is_reference<T>::value 
>> == true, void>::type* = nullptr>
>>      constexpr const char *modifier() const {
>>          return "[out] ";
>>      }
>> 
>> my attempt at it is like this:
>> 
>>    template modifier(T) {
>> 
>>        static if (is (T==const)) {
>> 
>>            const char* modifier = "[in] ";
>> 
>>        } else static if (/* T is a reference ?*/) { // [*]
>> 
>>            const char* modifier = "[out] ";
>>        }
>>    }
>> 
>> but even if I could e.g. say something like
>>    is(T == ref R, R),
>>    auto a = modifier!(ref T);
>> wouldn't work.
>> 
>> 
>> 
>
> did you try https://dlang.org/spec/traits.html#isRef?

yes, I did read the spec. I read the language spec on traits as 
well as std.traits docs as well as searching the internet for a 
solution since day before yesterday. But I couldn't bring it 
together because

   } else static if (__traits(isRef, T)) {

compiles, but e.g.

    assert (modifier!(ref int) == "[out] ");

doesn't.
Anyways, thanks for your reply.