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
暂无找到可以解决该程序问题的有效方法,小编努力寻找整理中!
如果你已经找到好的解决方法,欢迎将解决方案带上本链接一起发送给小编。
小编邮箱:dio#foxmail.com (将#修改为@)