[英]Daisy chain variadic templated classes
我有一個類模板管道:
template <typename A, typename B> class Pipeline;
我想創建一個可變參數函數模板,以接受任意數量的任意Pipeline
,並且希望以一種棘手的方式來約束那些模板。 非可變參數代碼如下所示:
Pipeline<A, C> compose(Pipeline<A, B> p1, Pipeline<B, C> p2);
Pipeline<A, D> compose(Pipeline<A, B> p1, Pipeline<B, C> p2, Pipeline<C, D> p3);
// ...and so on
現在是否有可能以可變的方式端到端地約束它們?
// I would like to write something like:
Pipeline<Args[0], Args[len(Args)-1]> compose(Pipeline<Args[i], Args[i+1]> ps...);
我假設您的Pipeline
看起來像這樣:
template <typename A, typename B>
struct Pipeline {
using first = A;
using second = B;
};
首先,讓我們為有效的鏈鏈接建立類型特征:
template <typename P1, typename P2>
struct is_valid_link : std::false_type { };
template <typename A, typename B, typename C>
struct is_valid_link<Pipeline<A,B>, Pipeline<B,C>> : std::true_type { };
接下來,讓我們借用bool_pack
的bool_pack
技巧,以驗證一堆bool
都是true
:
template <bool...> struct bool_pack;
template <bool... v>
using all_true = std::is_same<bool_pack<true, v...>, bool_pack<v..., true>>;
當然,我們需要索引序列的技巧:
template <typename... Pipelines,
typename R = decltype(detail::daisy_chain(
std::make_index_sequence<sizeof...(Pipelines)-1>(),
std::declval<Pipelines>()...))
>
R compose(Pipelines... pipelines)
{
return {};
}
大部分工作在這里進行檢查:
namespace detail {
template <size_t... Is,
typename... Pipelines,
typename T = std::tuple<Pipelines...>,
typename R = std::enable_if_t<
// ensure that all our pairwise pipelines are valid links
all_true<
is_valid_link<std::tuple_element_t<Is,T>,
std::tuple_element_t<Is+1,T>>::value...
>::value,
// pick out the first and last types
Pipeline<typename std::tuple_element_t<0, T>::first,
typename std::tuple_element_t<sizeof...(Pipelines)-1, T>::second>
>>
R daisy_chain(std::index_sequence<Is...>, Pipelines... pipelines);
}
這讓我們可以:
int main() {
Pipeline<int, double> p = compose(Pipeline<int, char>{}, Pipeline<char, double>{});
}
用這種方式編寫的好處是您仍然擁有SFINAE-如果您想要這樣的話。 這樣:
auto invalid = compose(Pipeline<int, char>{}, Pipeline<float, double>{});
將觸發重載解析失敗:
main.cpp:46:31: error: no matching function for call to 'compose'
Pipeline<int, double> p = compose(Pipeline<int, char>{}, Pipeline<float, double>{});
^~~~~~~
main.cpp:40:3: note: candidate template ignored: substitution failure [with Pipelines = <Pipeline<int, char>, Pipeline<float, double>>]: no matching function for call to 'daisy_chain'
R compose(Pipelines... pipelines)
^
1 error generated.
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