Pass C++11 enum class as template while auto deducing its type

Question

This minimal example compiles without warnings and runs:

// library
template<class T, T t> struct library_struct {};

// user
enum class my_enum { x, y, z };
int main()
{
    library_struct<my_enum, my_enum::x> unused; // l.7
    (void) unused;
    return 0;
}

Now, I want the compiler to deduce the type template parameter my_enum from the enum template paramter my_enum::x. This would look much nicer:

library_struct<my_enum::x> unused;

I have seen examples where the compiler was able to deduce template parameters, but I was only allowed to omit the last template parameters in the template parameter lists. So is it possible to omit the enum type here?

EDIT: I'm interested in solutions without macros.

Answer 1

There are 3 approaches, none of them good.

First, you could wait for a later standard: a number of proposals to fix this problem have been made. I do not know if any made it into C++1y.

Second, macros.

Third, use a deduced type. This forces the enum value to be at best a constexpr parameter.

The shorter answer is 'you cannot do what you ask, at least not cleanly'. The mess has been noted, and may one day be fixed.