枚舉的 C++ 模板特化

Question

我想將已知類型映射到我自己定義的枚舉值。

enum class MyType : uint8_t {
    Int8,
    Uint8,
    Int16,
    Uint16,
    Int32,
    Uint32,
    ... // some other primitive types.
};

template <typename T>
constexpr uint8_t DeclTypeTrait();

template <>
constexpr uint8_t DeclTypeTrait<int8_t>() {
  return static_cast<uint8_t>(MyType::Int8);
}
... // Specialize for each known number type.

同樣對於用戶定義的任何枚舉類型，我想將其映射到 Int32。 用戶必須在 int32_t 上定義他的枚舉類。

using Enumeration = int32_t;

// Some user defined enum.
enum class CameraKind : Enumeration { 
    Perspective, 
    Orthographic 
};

所以我像這樣實現 DeclTypeTrait ：

template <typename T,
          class = typename std::enable_if<std::is_enum<T>::value>::type>
constexpr uint8_t DeclTypeTrait() {
  return static_cast<uint8_t>(MyType::Int32);
}

但我收到錯誤：“調用 'DeclTypeTrait' 不明確”

candidate function [with T = CameraKind]
candidate function [with T = CameraKind, $1 = void]

我的問題是如何做到這一點：

// If a have a variable of known types or any enum on int32.
int8_t v1;
CameraKind camera;
std::string s;

DeclTypeTrait<decltype(v1)>() -> MyType::Int8
DeclTypeTrait<decltype(camera)>() -> MyType::Int32
DeclTypeTrait<decltype(s)>() // Report compile error is OK.

Answer 1

對於您的情況，使用類模板會簡單得多。

template <typename T, typename = std::void_t<>>
struct DeclTypeTraitT {
};

template <typename T>
inline constexpr uint8_t DeclTypeTrait = DeclTypeTraitT<T>::value;

template <>
struct DeclTypeTraitT<int8_t> {
    static constexpr uint8_t value = static_cast<uint8_t>(MyType::Int8);
};

template <typename T>
struct DeclTypeTraitT<T, std::enable_if_t<std::is_enum_v<T>>> {
    static constexpr uint8_t value = static_cast<uint8_t>(MyType::Int32);
};

然后

CameraKind camera;
static_assert(DeclTypeTrait<decltype(camera)> == static_cast<uint8_t>(MyType::Int32));

演示

---

正如@HolyBlackCat 所說，如果你想映射底層類型，你可以使用

template <typename T>
struct DeclTypeTraitT<T, std::enable_if_t<std::is_enum_v<T>>> {
    static constexpr uint8_t value = DeclTypeTrait<std::underlying_type_t<T>>;
};