构造函数的类模板参数包扩展

Question

I'd like to make a class template RestrictedInteger that can only be constructed with certain values known at compile time. This is how I could do it manually:

// Wrapper
template<int... Is> using IntList = std::integer_sequence<int, Is...>;

// This is my class
template<class intList> class RestrictedInteger;
template<int I1>
class RestrictedInteger<IntList<I1>> {
  const int _i;
public:
  constexpr RestrictedInteger(std::integral_constant<int, I1>) : _i(I1) {}
};
//[...]
template<int I1, I2, I3>
class RestrictedInteger<IntList<I1, I2, I3>> {
  const int _i;
public:
  constexpr RestrictedInteger(std::integral_constant<int, I1>) : _i(I1) {}
  constexpr RestrictedInteger(std::integral_constant<int, I2>) : _i(I2) {}
  constexpr RestrictedInteger(std::integral_constant<int, I3>) : _i(I3) {}
};
//[...] (and so on)

Naturally, I'd like to use a variadic template instead. If only this were legal :

template<int... Is>
class RestrictedInteger<IntList<Is...>> {
  int _i;
public:
  constexpr RestrictedInteger(std::integral_constant<int, Is>) : _i(Is) {}... // ERROR
}

Since I'm using C++17 however, I thought it would work like this:

template<int... Is>
class RestrictedInteger<IntList<Is...>> {
  int _i;
public:
  template<int I>
  constexpr RestrictedInteger(std::enable_if_t<...||(I==Is), std::integral_constant<int, I>>) : _i(I) {} // syntax error: '...' (Visual Stuio 2019)
};

But apparently not.

Any ideas of a neat way to solve this?

Answer 1

If failing compilation is an option (you don't need compiler to find other overloads) - you can put static_assert inside your constructor:

#include <type_traits>
#include <utility>

template<int... Is> using IntList = std::integer_sequence<int, Is...>;

template<class intList> class RestrictedInteger;

template<int... Is>
class RestrictedInteger<IntList<Is...>> {
private:
  const int _i;
public:
  template <int I>
  constexpr RestrictedInteger(std::integral_constant<int, I>) : _i(I) 
  {
      static_assert(((I == Is) || ...), "Invalid value");
  }
};

int main()
{
    RestrictedInteger<IntList<1, 2, 3>> i = std::integral_constant<int, 3>();
    RestrictedInteger<IntList<1, 2, 3>> ii = std::integral_constant<int, 6>(); // fails
}

or a bit more verbose solution with std::enable_if :

#include <type_traits>
#include <utility>

template<int... Is> using IntList = std::integer_sequence<int, Is...>;

template<class intList> class RestrictedInteger;

template<int... Is>
class RestrictedInteger<IntList<Is...>> {
private:
  const int _i;
public:
  template <int I, typename std::enable_if_t<((I == Is) || ...)>* = nullptr>
  constexpr RestrictedInteger(std::integral_constant<int, I>) : _i(I) 
  {
  }
};

int main()
{
    RestrictedInteger<IntList<1, 2, 3>> i = std::integral_constant<int, 3>();
    RestrictedInteger<IntList<1, 2, 3>> ii = std::integral_constant<int, 6>(); // fails
}