问题描述
是否有可能(如果可以,如何)从索引的特征类型集合中生成模板包,以便将其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;
解决方法
您可以这样做:
#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,要以字符串文字形式找到类型名称,可以只使用typeid(TYPE).name()
。如果需要,您可能需要取消使用此名称的名称;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,但是在GCC上,您可以在abi::__cxa_demangle
标头中使用<cxxabi.h>
。)