我的可变参数模板构造函数隐藏了复制构造函数，防止 class 被复制

Question

I made a Vector<numType, numberOfCoords> class. I was happy with it, it's kinda weird but it seemed to work at the start. But I just found out copying the vectors is impossible.

The reason is that to allow the number of coordinates to be a template, there is a templated variadic constructor which expects the right number of coordinates.

This is what it looks like, with the utility math methods removed:

template <typename NumType, unsigned char Size>
class Vector
{
public:
  using CoordType = NumType;

  //Vector(const Vector& v) : values(v.values) {}
  //Vector(Vector&& v) : values(std::move(v.values)) {}

  template<typename... NumTypes>
  constexpr Vector(NumTypes&&... vals) : values{ std::forward<NumTypes>(vals)... }
  {
    static_assert(sizeof...(NumTypes) == Size, "You must provide N arguments.");
  }

  Vector(const std::array<NumType, Size>& values) : values(values) {}
  Vector(std::array<NumType, Size>&& values) : values(std::move(values)) {}

  const NumType& operator[](size_t offset) const { return values[offset]; }
  NumType& operator[](size_t offset) { return values[offset]; }
  //Vector& operator=(const Vector& other) { values = other.values; }
  //Vector& operator=(Vector&& other) { values = std::move(other.values); }

  std::array<NumType, Size> values;

};

As you can see - now commented out - I did try to implement copy and move and assignment operations manually. It had no effect, the error is always along the lines of:

cannot convert ‘Vector<int, 3>’ to ‘int’ in initialization

This is because it is still trying to use the variadic constructor, instead of the copy constructor.

Here is a full working broken sample: https://ideone.com/Jz86vP

Answer 1

You might:

• SFINAE your variadic constructor, for example:

   template <typename... NumTypes, std::enable_if_t<sizeof...(NumTypes) == Size && std::conjunction<std::is_same<NumType, NumTypes>::value, int> = 0> constexpr Vector(NumTypes&&... vals): values{ std::forward<NumTypes>(vals)... } { }

  Note: std::conjunction is C++17, but can be done in C++14.

• or prepend a tag:

   template <typename... NumTypes> constexpr Vector(struct SomeTag, NumTypes&&... vals): values{ std::forward<NumTypes>(vals)... } { }

• or redesign the class, for example, something like:

   template <typename T, std::size_t> using always_type = T; template <typename NumType, typename Seq> class VectorImpl; template <typename NumType, std::size_t... Is> class VectorImpl<NumType, std::index_sequence<Is...>> { public: using CoordType = NumType; VectorImpl(const VectorImpl&) = default; VectorImpl(VectorImpl&&) = default; VectorImpl& operator=(const VectorImpl&) = default; VectorImpl& operator=(VectorImpl&&) = default; constexpr VectorImpl(always_type<NumType, Is>... vals): values{ std::forward<NumType>(vals)... } { } VectorImpl(const std::array<NumType, sizeof...(Is)>& values): values(values) {} VectorImpl(std::array<NumType, sizeof...(Is)>&& values): values(std::move(values)) {} const NumType& operator[](size_t offset) const { return values[offset]; } NumType& operator[](size_t offset) { return values[offset]; } std::array<NumType, sizeof...(Is)> values; }; template <typename T, std::size_t N> using Vector = VectorImpl<T, std::make_index_sequence<N>>;

  Demo