从一组特征创建模板包 [英] Create template pack from set of traits

问题描述

是否可以(如果可以，如何)从一组索引的类型特征生成模板包，以便它们可以用于实例化变体或元组?

Is it possible (and if so, how) to generate a template pack from a indexed set of type traits so they can be used to instantiate a variant or a tuple?

#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:

Or is it necessary to use the variant as input instead:

#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
};

或者甚至推出你自己的variant，它接受一组特征而不是一个简单的类型列表:

or even roll your own variant, which accepts a set of traits instead of a plain list of type:

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

推荐答案

您可以:

#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;

注意要以字符串文字形式查找类型名称，您可以使用 typeid(TYPE).name().如果您愿意，您可能需要修改此名称；您可以使用特定于编译器的 demangler 函数(我认为 MSVC 不会修改类型名称，但是在 GCC 上，您可以在 <cxxabi.h> 中使用 abi::__cxa_demangle> 标题.)

Note to find the typename as a string literal, you could just use typeid(TYPE).name(). You may need to demangle this name, if you wish; you can use your compiler-specific demangler function (I think MSVC doesn't mangle type names, but on GCC you would use abi::__cxa_demangle in <cxxabi.h> header.)

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