16

我实际上是想看看我是否可以获得一个支持我从 boost::fusion 中使用的极少数操作的最小库。

这是我到目前为止... 

template < typename... Types >
struct typelist
{
};

template < template < typename... > class F, typename... Args >
struct apply
{
  typedef typename F < Args... >::type type;
};

template < typename, template < typename... > class >
struct foreach;

template < typename... Types, template < typename Arg > class F >
struct foreach < typelist < Types... >, F >
{
  typedef typelist < typename apply < F, Types >::type... > type; 
};

由于元功能foreach实现很简单，我认为zip也很容易。显然，事实并非如此。

template < typename... >
struct zip;

template < typename...  Types0, typename... Types1 >
struct zip < typelist < Types0... >, typelist < Types1... > >
{
  typedef typelist < typelist < Types0, Types1 >... > type;
};

如何将此zip元函数推广到任意数量的类型列表？我们这里需要的似乎是参数包的参数包。我不知道该怎么做。

编辑1：

实施is_equal...

template < std::size_t... Nn >
struct is_equal;

template < std::size_t N0, std::size_t N1, std::size_t... Nn >
struct is_equal < N0, N1, Nn... >
: and_ <
    typename is_equal < N0, N1 >::type
  , typename is_equal < N1, Nn... >::type
  >::type
{
};

template < std::size_t M, std::size_t N >
struct is_equal < M, N > : std::false_type
{
  typedef std::false_type type;
};

template < std::size_t N >
struct is_equal < N, N > : std::true_type
{
  typedef std::true_type type;
};

zip我认为也可以采取类似的方法……还没有尝试过zip，但是当我回到家时会这样做。

编辑2：

这是我最终认为看起来更优雅的东西。这基本上是 Vaughn Cato 方法的一种变体。

namespace impl
{

template < typename Initial, template < typename, typename > class F, typename... Types >
struct foldl;

template < typename Initial, template < typename, typename > class F, typename First, typename... Rest >
struct foldl < Initial, F, First, Rest... >
{
  typedef typename foldl < typename F < Initial, First >::type, F, Rest... >::type type;
};

template < typename Final, template < typename, typename > class F >
struct foldl < Final, F >
{
  typedef Final type;
};

template < typename Type, typename TypeList >
struct cons;

template < typename Type, typename... Types >
struct cons < Type, typelist < Types... > >
{
  typedef typelist < Types..., Type > type;
};

template < typename, typename >
struct zip_accumulator;

template < typename... Types0, typename... Types1 >
struct zip_accumulator < typelist < Types0... >, typelist < Types1... > >
{
  typedef typelist < typename cons < Types1, Types0 >::type... > type;
};

template < typename... Types0 >
struct zip_accumulator < typelist <>, typelist < Types0... > >
{
  typedef typelist < typelist < Types0 >... > type;
};

template < typename... TypeLists >
struct zip
{
  typedef typename foldl < typelist <>, zip_accumulator, TypeLists... >::type type;
};

}

template < typename... TypeLists >
struct zip
{
  static_assert(and_ < typename is_type_list < TypeLists >... >::value, "All parameters must be type lists for zip");
  static_assert(is_equal < TypeLists::length... >::value, "Length of all parameter type lists must be same for zip");
  typedef typename impl::zip < TypeLists... >::type type;
};

template < typename... TypeLists >
struct zip < typelist < TypeLists... > > : zip < TypeLists... >
{
};

这被视为zip一个fold操作。

4

2 回答 2

6

这是我发现的最短的实现：

template <typename...> struct typelist { };   
template <typename A,typename B> struct prepend;
template <typename A,typename B> struct joincols;
template <typename...> struct zip;    

template <typename A,typename... B>
struct prepend<A,typelist<B...> > {
  typedef typelist<A,B...> type;
};

template <>
struct joincols<typelist<>,typelist<> > {
  typedef typelist<> type;
};

template <typename A,typename... B>
struct joincols<typelist<A,B...>,typelist<> > {
  typedef typename
    prepend<
      typelist<A>,
      typename joincols<typelist<B...>,typelist<> >::type
    >::type type;
};

template <typename A,typename... B,typename C,typename... D>
struct joincols<typelist<A,B...>,typelist<C,D...> > {
  typedef typename
    prepend<
      typename prepend<A,C>::type,
      typename joincols<typelist<B...>,typelist<D...> >::type
    >::type type;
};

template <>
struct zip<> {
  typedef typelist<> type;
};

template <typename A,typename... B>
struct zip<A,B...> {
  typedef typename joincols<A,typename zip<B...>::type>::type type;
};
于 2011-09-25T05:22:00.673 回答
3

似乎可以使用成熟的列表（这意味着 head、tail 和 cons 操作）和递归。用 GCC 4.7 的快照测试，所有的std东西都来自<type_traits>：

struct nil {};

template<typename T>
struct is_nil: std::is_same<T, nil> {};

template<typename... T>
struct and_: std::true_type {};

template<typename First, typename... Rest>
struct and_<First, Rest...>
: std::integral_constant<
    bool
    , First::value && and_<Rest...>::value
> {};

template<typename T>
struct not_
: std::integral_constant<bool, !T::value> {};

template<typename... T>
struct typelist;

template<typename First, typename Second, typename... Rest>
struct typelist<First, Second, Rest...> {
    typedef First head;
    typedef typelist<Second, Rest...> tail;
};

template<typename Last>
struct typelist<Last> {
    typedef Last head;
    typedef nil tail;
};

template<typename T, typename List>
struct cons;

template<typename T, typename... Ts>
struct cons<T, typelist<Ts...>> {
    typedef typelist<T, Ts...> type;
};

// workaround for:
// sorry, unimplemented: cannot expand '...' into a fixed-length argument list
template<template<typename...> class Template, typename... T>
struct gcc_workaround {
    typedef Template<T...> type;
};

namespace detail {

template<typename Sfinae, typename... Lists>
struct zip;

template<typename... Lists>
struct zip<
    typename std::enable_if<and_<is_nil<typename Lists::tail>...>::value>::type
    , Lists...
> {
    typedef typelist<typelist<typename Lists::head...>> type;
};

template<typename... Lists>
struct zip<
    typename std::enable_if<and_<not_<is_nil<typename Lists::tail>>...>::value>::type
    , Lists...
> {
    typedef typename cons<
        typelist<typename Lists::head...>
        , typename gcc_workaround<zip, void, typename Lists::tail...>::type::type
    >::type type;
};

} // detail

template<typename... Lists>
struct zip: detail::zip<void, Lists...> {};

您可能希望为所有这些添加错误检查（我正在考虑当前只是保留为不完整类型的无效实例化）。坦率地说，考虑到我花了很多时间来解决这个问题，我建议坚持使用 Boost.MPL。像懒惰的评估（我不需要做 SFINAE 的东西）这样的事情是一个福音，我不喜欢重新发明它们。再加上它启用 C++11 的那一天，您将获得两全其美。

我忘了提 Boost.MPL 还具有通用性的优点。它可以在满足其序列概念之一的任何类型上工作（也可以非侵入性地适应预先存在的类型），而您强制使用typelist.

于 2011-09-24T16:18:15.047 回答