过滤参数包的类型 [英] Filter the types of a parameter pack
问题描述
我想知道是否可以过滤传递给可变参数模板的类型(基于谓词模板)以生成另一个可变参数模板，其中包含满足谓词的类型:
I'd like to know if it's possible to filter the types passed to a variadic template (based on a predicate template) to produce another variadic template containing those types which satisfy the predicate:
/** 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，我还没有找到一种从其余过滤类型中提取"参数包的方法.
As you can see, I haven't found a way to "extract" the parameter pack from the rest of the filtered types.
提前谢谢！
推荐答案
那应该很简单.内心应该有这样的东西:
That should be fairly straight-forward. At the heart you should have something like this:
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> ，这会将 T，std :: tuple< Args ...> 转换为 std :: tuple< T，Args ...> ，并且您需要沿途传递谓词(作为练习).缺点可能看起来像这样:
You just need Cons<T, Tuple>, which turns T, std::tuple<Args...> into std::tuple<T, Args...>, and you need to pass the predicate along (left as an exercise). Cons could look like this:
template <typename, typename> struct Cons;
template <typename T, typename ...Args>
struct Cons<T, std::tuple<Args...>>
{
using type = std::tuple<T, Args...>;
};
filter< Args ...> :: type 的结果将是 std :: tuple< Brgs ...> ，其中 Brgs... 是一个包，仅包含谓词所保存的 Args ... 中的那些类型.
The result of filter<Args...>::type would be std::tuple<Brgs...>, where Brgs... is a pack consisting of only those types in Args... for which the predicate holds.
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