Question

我有一个类模板Pipeline：

template <typename A, typename B> class Pipeline;

我想创建一个可变参数函数模板，接受任意数量的任意Pipeline s，我想以一种棘手的方式约束它们。非可变代码看起来像：

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

现在有可能以可变方式端到端地约束它们吗？

// I would like to write something like:
Pipeline<Args[0], Args[len(Args)-1]> compose(Pipeline<Args[i], Args[i+1]> ps...);

Answer 1

我假设您的Pipeline看起来像：

template <typename A, typename B>
struct Pipeline {
    using first = A;
    using second = B;
};

首先，让我们为什么是有效链链接做出类型特征：

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 { };

接下来，让我们借用@ Columbo的bool_pack技巧来验证一堆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>>;

当然我们需要索引序列技巧：

template <typename... Pipelines,
          typename R = decltype(detail::daisy_chain(
                                    std::make_index_sequence<sizeof...(Pipelines)-1>(),
                                    std::declval<Pipelines>()...))
         >
R compose(Pipelines... pipelines)
{
    return {};
}

大部分工作都在这里进行检查：

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);
}

这让我们做到了：

int main() {
    Pipeline<int, double> p = compose(Pipeline<int, char>{}, Pipeline<char, double>{});
}

以这种方式编写它的好处是你还有SFINAE - 如果这是你想要的东西。这样：

auto invalid = compose(Pipeline<int, char>{}, Pipeline<float, double>{});

将触发重载解析失败：

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.

菊花链可变模板类

1 个答案: