c++ - 如何支持 Variant 类型的隐式转换，例如从 int 到 unsigned long？

Question

我想拥有我的 Variant类型，C++17 的包装器 std::variant , 在适当的地方隐式转换类型。例如，char至 std::string , 或 int至 unsigned long .这是我的代码:

#include <variant>

using variant_t = std::variant<
        std::monostate,
        std::string, bool, std::int32_t,
        std::uint32_t, std::int64_t, std::uint64_t,
        float, double, char, unsigned char,
        std::vector<double>>;

class Variant : public variant_t {
public:

    using variant::variant;

    enum TypeId {
        EMPTY = 0, // std::monostate. Empty is default when variant instantiated with nothing
        STRING = 1,
        BOOL = 2,
        INT32 = 3,
        UINT32 = 4,
        INT64 = 5,
        UINT64 = 6,
        FLOAT = 7,
        DOUBLE = 8,
        CHAR = 9,
        UCHAR = 10,
        DOUBLEVECTOR = 11
    };

    TypeId type() const {
        return (Variant::TypeId) index();
    }

    template<class VariantType>
    VariantType get() const {
        return std::get<VariantType>(*this);
    }
};

我希望能够做的是:

TEST(VariantTests, HowToConvertIntToULongWithoutManualCast) {
    Variant v(11);
    ASSERT_EQ(v.type(), Variant::TypeId::INT32); // (pass, v is an int)
    unsigned long toUnsignedLong = v; // error
    long toLong = v; // error
    // and any other conversions from int that make sense
}

如何修改我的 Variant支持隐式类型转换？
编辑
根据评论，我还需要考虑不兼容的对以及兼容的对，例如以下将失败。

Variant v(12); // variant containing an int
std::string x = v; // should error, int to string incompatible

Answer 1

沿着这些路线的东西:

template <typename T>
struct Visitor {
  template <typename U>
  std::enable_if_t<std::is_convertible_v<U, T>, T> operator() (U&& val) {
      return val;
  }

  template <typename U>
  std::enable_if_t<!std::is_convertible_v<U, T>, T> operator() (U&& val) {
      throw std::bad_variant_access{};
  }
};

template<class VariantType>
VariantType get() const {
    return std::visit(Visitor<VariantType>{}, static_cast<const variant_t&>(*this));
}

Demo