Question

我正在创建Node的通用集合。每个Node的类型分别为Start和End。并且其中一个的End类型必须与下一个的Start类型相匹配。

如果我要列出集合中的每种类型，则构造函数将如下所示（对于四种类型）：

template <typename Start, typename End>
class Node {
};

template <typename A, typename B, typename C, typename D>
class Collection
{
public:
    Collection(Node<A, B> n1, Node<B, C> n2, Node<C, D> n3) { }
};

但是当我尝试将constuctor编写为可变参数模板以支持任意数量的类型时，我感到很困惑。

Answer 1

我提出了一些不同的解决方案。

给出一个普通的tag结构来包装一个泛型类型（以避免在std::tuples s中无法默认构造的类型的问题）

template <typename>
struct tag
 { };

和一个基于std::tuple

定义2种类型的助手结构

template <typename...>
struct getTpls;

template <std::size_t ... Is, typename ... Ts>
struct getTpls<std::index_sequence<Is...>, Ts...>
 {
   using tpl0 = std::tuple<tag<Ts>...>;
   using ftpl = std::tuple<std::tuple_element_t<Is,    tpl0>...>;
   using stpl = std::tuple<std::tuple_element_t<1u+Is, tpl0>...>;
 };

您可以如下编写Collection

template <typename ... Ts>
struct Collection
 {
   static_assert( sizeof...(Ts) > 1u, "more types, please");

   using getT = getTpls<std::make_index_sequence<sizeof...(Ts)-1u>, Ts...>;

   using ftpl = typename getT::ftpl;
   using stpl = typename getT::stpl;

   template <typename ... FTs, typename ... STs,
             std::enable_if_t<
                 std::is_same_v<ftpl, std::tuple<tag<FTs>...>>
              && std::is_same_v<stpl, std::tuple<tag<STs>...>>, int> = 0>
   Collection (Node<FTs, STs> ...)
    { }
 };

以下是完整的编译示例

#include <tuple>
#include <type_traits>

template <typename Start, typename End>
class Node
 { };

struct A {};
struct B {};
struct C {};

template <typename>
struct tag
 { };

template <typename...>
struct getTpls;

template <std::size_t ... Is, typename ... Ts>
struct getTpls<std::index_sequence<Is...>, Ts...>
 {
   using tpl0 = std::tuple<tag<Ts>...>;
   using ftpl = std::tuple<std::tuple_element_t<Is,    tpl0>...>;
   using stpl = std::tuple<std::tuple_element_t<1u+Is, tpl0>...>;
 };

template <typename ... Ts>
struct Collection
 {
   static_assert( sizeof...(Ts) > 1u, "more types, please");

   using getT = getTpls<std::make_index_sequence<sizeof...(Ts)-1u>, Ts...>;

   using ftpl = typename getT::ftpl;
   using stpl = typename getT::stpl;

   template <typename ... FTs, typename ... STs,
             std::enable_if_t<
                 std::is_same_v<ftpl, std::tuple<tag<FTs>...>>
              && std::is_same_v<stpl, std::tuple<tag<STs>...>>, int> = 0>
   Collection (Node<FTs, STs> ...)
    { }
 };

int main ()
 {
   Collection<A, B, C>  c0{Node<A, B>{}, Node<B, C>{}};    // compile
   // Collection<A, B, B>  c1{Node<A, B>{}, Node<B, C>{}}; // error!
 }

Answer 2

通过一些间接的操作，您可以这样做：

template <typename Start, typename End>
class Node {
    // ...
};

// Implementation using the Nodes
// You might add typedef in Node to retrieve Start/End if needed (or create traits)
template <typename ... Nodes>
struct CollectionImpl
{
    CollectionImpl(Nodes ... ns) : nodes(ns...){}

    std::tuple<Nodes...> nodes; // You probably want something like that
};

// Helper class to build the type
template <typename Seq, typename Tup> struct CollectionMaker;

template <std::size_t ... Is, typename Tuple>
struct CollectionMaker<std::index_sequence<Is...>, Tuple>
{
    using type = CollectionImpl<Node<std::tuple_element_t<Is, Tuple>,
                                     std::tuple_element_t<Is + 1, Tuple>>...>;
};

// Wanted interface.
template <typename ... Ts>
using Collection = typename CollectionMaker<std::make_index_sequence<sizeof...(Ts) - 1>,
                                            std::tuple<Ts...>>::type;

Demo

Answer 3

基于max66的答案：

这将清理不必要的tag结构，并简化index_sequence来进行直接递归（类似于元组的定义）。

template <typename Owner, typename Value>
class Node {
};


struct A {};
struct B {};
struct C {};
struct D {};


template <typename First, typename...Rest>
std::tuple<First, Rest...> tuple_push_front(std::tuple<Rest...>);

template <typename T1, typename T2, typename...T>
struct NodeCollector {
private:
    using nodeRest = NodeCollector<T2, T...>;
public:
    using tplOwners = decltype(tuple_push_front<T1>(std::declval<typename nodeRest::tplOwners>()));
    using tplValues = decltype(tuple_push_front<T2>(std::declval<typename nodeRest::tplValues>()));
};
template <typename T1, typename T2>
struct NodeCollector<T1, T2> {
public:
    using tplOwners = std::tuple<T1>;
    using tplValues = std::tuple<T2>;
};


template <typename...Ts>
class Collection
{
    static_assert( sizeof...(Ts) > 1u, "Collection requires at least two types.");
private:
    using nodeCollector = NodeCollector<Ts...>;
public:
    template <typename...OTs, typename...VTs, typename=std::enable_if_t<
        (std::is_same_v<typename nodeCollector::tplOwners, std::tuple<OTs...>> &&
         std::is_same_v<typename nodeCollector::tplValues, std::tuple<VTs...>>)> >
    Collection(Node<OTs, VTs>...) { }
};


int main()
{
    Collection<A, B, C, D> c{Node<A, B>{}, Node<B, C>{}, Node<C, D>{}};
    std::cout << demangle(typeid(c).name()) << std::endl;
}

逐对展开可变参数模板包

3 个答案: