11

我有一个这样的构造：

template<typename... Ts>
struct List {}

typedef List<char,List<int,float,List<int,unsigned char>>,List<unsigned,short>> MyList;

我想基本上将它展平为一个列表。什么是最好的方法？我想如果我摆弄它的时间足够长，我可以用递归来做一些事情，但有些事情告诉我应该有更好的方法。

作为上述树的结果，我想要的应该类似于：

typedef List<char,int,float,int,unsigned char,unsigned,short> FlattenedList;

这是我的第一次尝试：

template<typename... Ts>
struct List{};

template<typename... Ts>
struct FlattenTree{
    typedef List<Ts...> Type;
};
template<typename... Ts, typename... Us, typename... Vs>
struct FlattenTree<Ts...,List<Us...>,Vs...>{
    typedef typename FlattenTree<Ts..., Us..., Vs...>::Type Type;
};

但这会导致此错误：error C3515: if an argument for a class template partial specialization is a pack expansion it shall be the last argument

rici 在这里指出了MSVC2013 抱怨的内容，所以这里没有编译器错误：

§ 14.8.2.5（从类型推导模板参数）第 5 段列出了不能推导模板参数的上下文。相关的是列表中的最后一个：

— A function parameter pack that does not occur at the end of the parameter-declaration-clause.

更新：

我想可以在最后放入一个虚拟参数，继续将第一个参数移动到末尾，或者如果它是 List 则将其扩展到前面，并专注于第一个参数是我的虚拟参数以停止递归。不过，对于编译器来说，为了展平列表似乎需要做很多工作。

namespace Detail{
    struct MyMagicType {};
    template<typename T, typename... Ts>
    struct FlattenTree{
        typedef typename FlattenTree<Ts..., T>::Type Type;
    };
    template<typename... Ts>
    struct FlattenTree<MyMagicType,Ts...>{      //termination case
        typedef List<Ts...> Type;
    };
    template<typename... Ts, typename... Us>
    struct FlattenTree<List<Ts...>, Us...>{
        typedef typename FlattenTree<Ts..., Us...>::Type Type;
    };              //expand Ts to front because they may hold more nested Lists
}

template<typename... Ts>
struct FlattenTree{
    typedef typename Detail::FlattenTree<Ts...,Detail::MyMagicType>::Type Type;
};

这适用于 MSVC2013，但我认为这不是最好的方法，因为我需要一个虚拟类型并且它给编译器带来了很多负载。我想将它与包含 500 多个元素的列表一起使用。

4

3 回答 3

3

另一种方法是使用辅助类和累加器列表而不是MyMagicType. 我们从一个空开始List<>，然后用输入列表中的类型填充它：

#include <type_traits>

template <class... Ts> struct List {};

// first parameter - accumulator
// second parameter - input list
template <class T, class U>
struct flatten_helper;

// first case - the head of the List is List too
// expand this List and continue
template <class... Ts, class... Heads, class... Tail>
struct flatten_helper<List<Ts...>, List<List<Heads...>, Tail...>> {
    using type = typename flatten_helper<List<Ts...>, List<Heads..., Tail...>>::type;
};

// second case - the head of the List is not a List
// append it to our new, flattened list
template <class... Ts, class Head, class... Tail>
struct flatten_helper<List<Ts...>, List<Head, Tail...>> {
    using type = typename flatten_helper<List<Ts..., Head>, List<Tail...>>::type;
};

// base case - input List is empty
// return our flattened list
template <class... Ts>
struct flatten_helper<List<Ts...>, List<>> {
    using type = List<Ts...>;
};

// wrapper around flatten_helper
template <class T> struct flatten;

// start with an empty accumulator
template <class... Ts>
struct flatten<List<Ts...>> {
    using type = typename flatten_helper<List<>, List<Ts...>>::type;
};

auto main() -> int {
    using Types = List<int, List<float, List<double, List<char>>>>;
    using Flat = flatten<Types>::type;

    static_assert(std::is_same<Flat, List<int, float, double, char>>::value, "Not the same");
}
于 2013-11-07T16:09:56.157 回答
2

您的解决方案非常优雅 IMO，这是我脑海中的另一个问题：

// the tuple-like class we want to flatten
// (i.e. the node of the tree, with several children as template parameters)
template<class... TT>
struct List
{};


// a join metafunction. Joins multiple Lists into a single List
// e.g.    List<TT1...>, List<TT2...>, etc., List<TTN...>
//      => List<TT1..., TT2..., etc., TTN...>
// requires: all template arguments are `List<..>`s
template<class... TT>
struct join
{
    using type = List<>; // end recursion for no parameters
};

template<class... TT>
struct join<List<TT...>>
{
    using type = List<TT...>; // end recursion for a single parameter
};

template<class... TT0, class... TT1, class... TT2>
struct join<List<TT0...>, List<TT1...>, TT2...>
{
    // join two adjacent lists into one, recurse
    // by joining the two lists `List<TT0...>` and `List<TT1...>`,
    // we get one template argument less for the next instantiation of `join`
    // this recurs until there's only one argument left, which then
    // matches the specialization `struct join<List<TT...>>`
    using type = typename join< List<TT0..., TT1...>, TT2... > :: type;
};


// the flatten metafunction
// guarantees (all specializations): the nested `type` is a flat `List<..>`
template<class T>
struct flatten
{
    // because of the partial specialization below,
    // this primary template is only used if `T` is not a `List<..>`
    using type = List<T>; // wrap the argument in a `List`
};

template<class... TT>
struct flatten<List<TT...>> // if the argument is a `List` of multiple elements
{
    // then flatten each element of the `List` argument
    // and join the resulting `List<..>`s
    using type = typename join<typename flatten<TT>::type...>::type;

    // ex. the argument is `List<List<int>, List<double>>`
    // then `TT...` is deduced to `List<int>, List<double>`
    // `List<int>` flattened is `List<int>`, similarly for `List<double>`
    // `join< List<int>, List<double> >` yields `List<int, double>`
};

使用和测试代码：

#include <iostream>
template<class T>
void print(T)
{
    std::cout << __PRETTY_FUNCTION__ << "\n"; // NON-STANDARD
}

int main()
{
    typedef List<char,List<int,float,List<int,unsigned char>>,
                 List<unsigned,short>> MyList;
    print( flatten<MyList>::type{} );
}

我确信最简单的方法是使用 boost::MPL ;)

于 2013-11-07T16:07:34.243 回答
0

这是我使用通用“元树遍历”的解决方案：

#include <iostream>
#include <type_traits>
#include <typeinfo>

template <typename T>
struct HasChildren : std::false_type {};

template <template <typename...> class P, typename... Types>
struct HasChildren<P<Types...>> : std::true_type {};

template <typename, typename> struct Visit;
template <typename, typename, bool> struct VisitHelper;
template <typename, typename> struct LeafAction;

// Here the role of P2<Visited...> is simply to allow LeafAction to carry out its function.  It is not native to the tree structure itself.
template <template <typename...> class P1, template <typename...> class P2, typename First, typename... Rest, typename... Visited>
struct Visit<P1<First, Rest...>, P2<Visited...>> :
    VisitHelper<P1<First, Rest...>, P2<Visited...>, HasChildren<First>::value> {};

template <template <typename...> class P1, template <typename...> class P2, typename... Visited>
struct Visit<P1<>, P2<Visited...>> {  // End of recursion.  Every leaf in the tree visited.
    using result = P2<Visited...>;
};

// Here First has children, so visit its children, after which visit Rest...
template <template <typename...> class P1, template <typename...> class P2, typename First, typename... Rest, typename... Visited>
struct VisitHelper<P1<First, Rest...>, P2<Visited...>, true> : Visit<P1<Rest...>, typename Visit<First, P2<Visited...>>::result> {};  // Visit the "subtree" First, and then after that visit Rest...  Need to use ::result so that when visiting Rest..., the latest value of the P2<Visited...> pack is used.

// Here First has no children, so the "leaf action" is carried out. 
template <template <typename...> class P1, template <typename...> class P2, typename First, typename... Rest, typename... Visited>
struct VisitHelper<P1<First, Rest...>, P2<Visited...>, false> : LeafAction<P1<First, Rest...>, P2<Visited...>> {};

// As a simple example, the leaf action shall simply be appending its type to P<Visited...>.
template <template <typename...> class P1, template <typename...> class P2, typename First, typename... Rest, typename... Visited>
struct LeafAction<P1<First, Rest...>, P2<Visited...>> : Visit<P1<Rest...>, P2<Visited..., First>> {};  // Having visited First, now visit Rest...

template <typename> struct VisitTree;

template <template <typename...> class P, typename... Types>
struct VisitTree<P<Types...>> : Visit<P<Types...>, P<>> {};

// ---------------------------- Testing ----------------------------
template <typename...> struct Pack;

template <typename Last>
struct Pack<Last> {
    static void print() {std::cout << typeid(Last).name() << std::endl;}
};

template <typename First, typename... Rest>
struct Pack<First, Rest...> {
    static void print() {std::cout << typeid(First).name() << ' ';  Pack<Rest...>::print();}
};

template <typename...> struct Group;
template <typename...> struct Wrap;
struct Object {};

int main() {
    VisitTree<
        Pack<Pack<int, Object, double>, bool, Pack<char, Pack<int, double, Pack<char, Pack<char, long, short>, int, Object>, char>, double>, long>
    >::result::print();  // int Object double bool char int double char char long short int Object char double long

    std::cout << std::boolalpha << std::is_same<
        VisitTree<Pack<Wrap<int, Object, double>, bool, Group<char, Pack<int, double, Pack<char, Wrap<char, long, short>, int, Object>, char>, double>, long>>::result,
        Pack<int, Object, double, bool, char, int, double, char, char, long, short, int, Object, char, double, long>
    >::value << std::endl;  // true
}
于 2015-04-04T18:34:54.883 回答