我想知道是否可以过滤传递给可变参数模板（基于谓词模板）的类型，以生成另一个包含满足谓词的类型的可变参数模板：
/** Filter a parameter pack */
template <template <class> class,
template <class...> class,
class...>
struct filter;
template <template <class> class Pred, template <class...> class Variadic>
struct filter<Pred, Variadic> : Variadic<>
{};
template <template <class> class Pred,
template <class...> class Variadic,
class T, class... Ts>
struct filter<Pred, Variadic, T, Ts...>
{
// FIXME: this just stops at first T where Pred<T> is true
using type = typename std::conditional<
Pred<T>::value,
Variadic<T, Ts...>, // can't do: Variadic<T, filter<...>>
filter<Pred, Variadic, Ts...> >::type;
};
如您所见，我还没有找到从其余过滤类型中“提取”参数包的方法。
提前致谢！
这应该是相当简单的。你内心应该有这样的东西：
template <typename...> struct filter;
template <> struct filter<> { using type = std::tuple<>; };
template <typename Head, typename ...Tail>
struct filter<Head, Tail...>
{
using type = typename std::conditional<Predicate<Head>::value,
typename Cons<Head, typename filter<Tail...>::type>::type,
typename filter<Tail...>::type
>::type;
};
您只需要
Cons<T, Tuple>T, std::tuple<Args...>std::tuple<T, Args...>Constemplate <typename, typename> struct Cons;
template <typename T, typename ...Args>
struct Cons<T, std::tuple<Args...>>
{
using type = std::tuple<T, Args...>;
};
filter<Args...>::typestd::tuple<Brgs...>Brgs...Args...