我又遇到了这个问题,想出了一个更通用的 Sebastian Redl 解决方案。
//given an index to replace at, a type to replace with and a tuple to replace in
//return a tuple of the same type as given, with the type at ReplaceAt set to ReplaceWith
template <size_t ReplaceAt, typename ReplaceWith, size_t... Idxs, typename... Args>
auto replace_type (std::index_sequence<Idxs...>, std::tuple<Args...>)
-> std::tuple<std::conditional_t<ReplaceAt==Idxs, ReplaceWith, Args>...>;
//instantiates a template with the types held in a tuple
template <template <typename...> class T, typename Tuple>
struct type_from_tuple;
template <template <typename...> class T, typename... Ts>
struct type_from_tuple<T, std::tuple<Ts...>>
{
using type = T<Ts...>;
};
//replaces the type used in a template instantiation of In at index ReplateAt with the type ReplaceWith
template <size_t ReplaceAt, typename ReplaceWith, class In>
struct with_n;
template <size_t At, typename With, template <typename...> class In, typename... InArgs>
struct with_n<At, With, In<InArgs...>>
{
using tuple_type = decltype(replace_type<At,With>
(std::index_sequence_for<InArgs...>{}, std::tuple<InArgs...>{}));
using type = typename type_from_tuple<In,tuple_type>::type;
};
//convenience alias
template <size_t ReplaceAt, typename ReplaceWith, class In>
using with_n_t = typename with_n<ReplaceAt, ReplaceWith, In>::type;
优点:
- 灵活选择要更改的参数
- 不需要改变原来的类
- 支持有一些参数没有默认值的类
-
options<int,long,int> 和 with_n_t<2,int,options<>> 是同一类型
一些使用示例:
with_n_t<1, int, options<>> a; //options<int, int, std::string>
with_n_t<2, int,
with_n_t<1, int, options<>>> b; //options<int, int, int>
您可以进一步将其概括为采用可变参数对的索引和类型,这样您就不需要嵌套with_n_t。