我实际上是想看看是否可以得到一个最小的库来支持我从 boost::fusion 中使用的极少数操作。

这是我目前所拥有的......

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; 
};

由于元函数 foreach 的实现很简单，我认为 zip 也很简单。显然，情况并非如此。

template < typename... >
struct zip;

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

如何将这个 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:

is_equal的实现...

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;
};

我认为zip 也可以采用类似的方法...尚未尝试使用zip，但我回家后会这样做。

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... >
{
};

这会将 zip 视为 fold 操作。

最佳答案

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;
};

关于c++ - 在 C++11 中实现元函数 zip，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7539716/