enable_if in template Parameters Creates Template

Question

In this answer what I really wanted to do is define a typename in my template parameters which could be used in the cast and return.

So this:

template <typename T>
typename std::enable_if<sizeof(unsigned char) == sizeof(T), unsigned char>::type caster(T value){ return reinterpret_cast<unsigned char&>(value); }

Would become this:

template <typename T, typename R = std::enable_if<sizeof(unsigned char) == sizeof(T), unsigned char>::type >
R caster(T value){ return reinterpret_cast<R&>(value); }

This works and behaves as desired for a single template specialization, but say that I add another specialization:

template <typename T, typename R = std::enable_if<sizeof(short) == sizeof(T), short>::type>
R caster(T value){ return reinterpret_cast<R&>(value); }

Now I get an error:

error C2995: 'R caster(T)' : function template has already been defined

Is there a way to convince the compiler that only one of these specializations will actually build for any given call?

Answer 1

No, there isn't. Template default arguments are just that, defaults. Any user could call caster<short, short>, which would match both overloads.

However, it would be possible to add more dummy arguments.

template <typename T,
          typename R = typename std::enable_if<sizeof(unsigned char) == sizeof(T), unsigned char>::type >
R caster(T value) { return reinterpret_cast<R&>(value); }

template <typename T,
          typename R = typename std::enable_if<sizeof(short) == sizeof(T), short>::type,
          typename = void>
R caster(T value) { return reinterpret_cast<R&>(value); }

(Also note the added typename.)

---

However, since all bodies are identical, I probably wouldn't go with overloads.

template <std::size_t N>
struct cast_result;

template <>
struct cast_result<sizeof(std::uint8_t)> {
  typedef std::uint8_t type;
};

template <>
struct cast_result<sizeof(std::uint16_t)> {
  typedef std::uint16_t type;
};

...

template <typename T, typename R = typename cast_result<sizeof(T)>::type>
R caster(T value) {
  return reinterpret_cast<R&>(value);
}

A final note: this use of reinterpret_cast is a violation of the aliasing rules. However, that's easily fixed:

template <typename T, typename R = typename cast_result<sizeof(T)>::type>
R caster(T value) {
  R result;
  std::memcpy(&result, &value, sizeof result);
  return result;
}

Answer 2

It seems that the best solution here may be to use a slew of conditionals, which would prevent me from having to fool with template specializations:

template <typename T, typename R = std::conditional<sizeof(T) == sizeof(unsigned char),
                                                    unsigned char,
                                                    conditional<sizeof(T) == sizeof(unsigned short),
                                                                unsigned short,
                                                                conditional<sizeof(T) == sizeof(unsigned long),
                                                                            unsigned long,
                                                                            enable_if<sizeof(T) == sizeof(unsigned long long), unsigned long long>::type>::type>::type>::type>
R caster(T value){ return reinterpret_cast<R&>(value); }

My apologies to the reader cause it's like reading nested ternaries. However I'm currently unaware of a cleaner way to handle this.

This sadly still doesn't prevent the user from stomping on all my defaulting by providing his own second template parameter as mentioned by hvd.

EDIT:

I've asked another question here which has a solution that doesn't require placing the typename in the template definition and doesn't require declaring the type twice.