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是否有可能（如果有，如何）从索引的类型特征集生成模板包，以便它们可用于实例化变体或元组？

#include <variant>

template<int n>
struct IntToType;

template<>
struct IntToType<0>
{
    using type = int;
    static constexpr char const* name = "int";
//  Other compile-time metadata
};

template<>
struct IntToType<1>
{
    using type = double;
    static constexpr char const* name = "double";
//  Other compile-time metadata
};

using MyVariant = std::variant<IntToType<???>::type...>;  // something with make_integer_sequence and fold expression?

或者是否有必要使用变体作为输入：

#include <variant>

using MyVariant = std::variant<int, double>;

template<int n>
struct IntToTypeBase
{
    using type = std::variant_alternative_t<n, MyVariant>;
};

template<int >
struct IntToType;

template<>
struct IntToType<0>:IntToTypeBase<0>
{
    static constexpr char const* name = "int";
//  Other compile-time metadata
};

template<>
struct IntToType<1>:IntToTypeBase<1>
{
    static constexpr char const* name = "double";
//  Other compile-time metadata
};

甚至滚动你自己的variant，它接受一组特征而不是简单的类型列表：

template<class IntegerType, template<auto> class Traits, size_t LastIndex>
class Variant;
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1 回答 1

2

你可以这样做：

#include <variant>

template<int n>
struct IntToType;

template<>
struct IntToType<0>
{
    using type = int;
    static constexpr char const* name = "int";
//  Other compile-time metadata
};

template<>
struct IntToType<1>
{
    using type = double;
    static constexpr char const* name = "double";
//  Other compile-time metadata
};

// replace NUMBER_OF_TYPES
template <typename T=std::make_index_sequence<NUMBER_OF_TYPES> >
struct make_my_variant;

template <size_t... indices>
struct make_my_variant<std::index_sequence<indices...> > {
    using type = std::variant<typename IntToType<indices>::type...>;
};

using MyVariant = typename std::make_my_variant<>::type;

请注意，要将 typename 查找为字符串文字，您可以只使用typeid(TYPE).name(). 如果您愿意，您可能需要解除此名称；您可以使用特定于编译器的 demangler 函数（我认为 MSVC 不会破坏类型名称，但在 GCC 上您可以abi::__cxa_demangle在<cxxabi.h>标头中使用。）

于 2020-10-23T18:24:31.927 回答