10

出于愚蠢的原因，我不会在这里讨论，我需要注释掉的行才能工作，而它上面的行不能工作：

template<uint _N, typename... _Args>
struct PartialTuple;

template<uint _N, typename _Arg, typename... _Args>
struct PartialTuple<_N, _Arg, _Args...>: PartialTuple<_N-1, _Args...> {};

template<typename _Arg, typename... _Args>
struct PartialTuple<0, _Arg, _Args...>
{
    typedef std::tuple<_Arg, _Args...> type;
};

int main()
{
    // I want this to not work...
    PartialTuple<1, std::string, std::string, int, int>::type A{"test", 5, 1};

    // I want this to work...
    //PartialTuple<1, std::string, std::string, int, int>::type B{"test", "test", 5};
}

我尝试_Arg与交换_Args...，但这不会编译（至少在 GCC 4.6 中）：

error: parameter pack argument ‘_Args ...’ must be at the end of the template argument list

如何从尾部而不是头部拉出物品？

4

4 回答 4

5

这是一个解决方案：N我没有从后面截断，而是sizeof...(Args) - N从前面截断：

#include <tuple>

/* Concatenator helper */

template <typename T, typename Tuple> struct cat;
template <typename T, typename ...Args>
struct cat<T, std::tuple<Args...>>
{
  typedef typename std::tuple<T, Args...> value;
};


/* Head-of-tuple */

template <unsigned int, typename...> struct tuple_head;

// Base case. Need to specialize twice, once for one and once for variadic types
template <typename ...Args>
struct tuple_head<0, Args...>
{
  typedef std::tuple<> value;
};
template <typename T>
struct tuple_head<0, T>
{
  typedef std::tuple<> value;
};

// Recursion step
template <unsigned int N, typename T, typename ...Args>
struct tuple_head<N, T, Args...>
{
  typedef typename cat<T, typename tuple_head<N - 1, Args...>::value>::value value;
};


/* User interface */

template <unsigned int N, typename ...Args>
struct PartialTuple
{
  typedef typename tuple_head<sizeof...(Args) - N, Args...>::value type;
};


/* Usage */

#include <string>
int main()
{
  // I want this to not work...
  //PartialTuple<1, std::string, std::string, int, int>::type A{"test", 5, 1};

  // I want this to work...
  PartialTuple<1, std::string, std::string, int, int>::type B("test", "test", 5);
  PartialTuple<0, std::string, std::string, int, int>::type C("test", "test", 5, 6);
}
于 2011-07-09T14:54:04.293 回答
3

我整晚都在玩它，终于找到了一些工作（改变了我的外壳以匹配 STL）：

template<uint _N, typename... _All>
struct reverse_tuple_outer
{
    template<typename _Head, typename... _Tail>
    struct reverse_tuple_inner: reverse_tuple_outer<_N-1, _Head, _All...>::template reverse_tuple_inner<_Tail...> { };
};

template<typename... _All>
struct reverse_tuple_outer<0, _All...>
{
    template<typename... _Tail>
    struct reverse_tuple_inner {
        typedef std::tuple<_All...> type;
    };
};

template<typename... _Args>
struct reverse_tuple
{
    typedef typename reverse_tuple_outer<sizeof...(_Args)>::template reverse_tuple_inner<_Args...>::type type;
};

template<typename... _Args>
struct strip_and_reverse_tuple;

template<typename... _Args>
struct strip_and_reverse_tuple<std::tuple<_Args...>>
{
    typedef typename reverse_tuple<_Args...>::type type;
};

template<uint _N, typename... _Args>
struct partial_tuple
{
    typedef typename strip_and_reverse_tuple<typename reverse_tuple_outer<sizeof...(_Args)-_N>::template reverse_tuple_inner<_Args...>::type>::type type;
};

int main()
{
    //partial_tuple<1, std::string, std::string, int, int>::type A{"test", 5, 1};
    partial_tuple<1, std::string, std::string, int, int>::type B{"test", "test", 5};
}

作为额外的奖励，我也有reverse_tuple，如果我需要它。

于 2011-07-09T08:14:59.737 回答
3

我让我的代码工作起来有点像 Haskell 中的列表——因为，嗯，TMP 是 C++ 中的纯函数式语言。

add_to_pack相当于 Haskell 的列表构造函数(:)。drop_from_end实现为（在 Haskell 表示法中）\x list -> take (length list - x) list，其中take n只取n列表的第一个元素。

我想你可以std::tuple直接使用而不是pack，但我更喜欢这个解决方案，因为它不会滥用元组作为模板参数包持有者。:)

这是代码：

#include <tuple>
#include <type_traits> // for std::conditional


template <typename... Pack>
struct pack
{ };


template <typename, typename>
struct add_to_pack;

template <typename A, typename... R>
struct add_to_pack<A, pack<R...>>
{
  typedef pack<A, R...> type;
};


template <typename>
struct convert_to_tuple;

template <typename... A>
struct convert_to_tuple<pack<A...>>
{
  typedef std::tuple<A...> type;
};


template <int, typename...>
struct take;

template <int N>
struct take<N>
{
  typedef pack<> type;
};

template <int N, typename Head, typename... Tail>
struct take<N, Head, Tail...>
{
  typedef
    typename std::conditional<
      (N > 0),
      typename add_to_pack<
        Head,
        typename take<
          N - 1,
          Tail...
        >::type
      >::type,
      pack<>
    >::type type;
};  


template <int N, typename... A>
struct drop_from_end
{
  // Add these asserts if needed.
  //static_assert(N >= 0,
  //  "Cannot drop negative number of elements!");

  //static_assert(N <= static_cast<int>(sizeof...(A)),
  //  "Cannot drop more elements than size of pack!")

  typedef
    typename convert_to_tuple<
      typename take<
        static_cast<int>(sizeof...(A)) - N,
        A...
      >::type
    >::type type;
};


int main()
{
  drop_from_end<2, const char*, double, int, int>::type b{"pi", 3.1415};
}

这是工作代码：通过 ideone.com。

该take结构或多或少等同于以下 Haskell 代码：

take n []     = []
take n (x:xs)
  | n > 0     = x : take (n - 1) xs
  | otherwise = []
于 2011-07-09T14:35:14.753 回答
0

我使用 Boost.MPL 和 Boost.Fusion 做了类似的事情：使用 MPL 工具计算类型序列，例如push_back，然后将其转换为fusion::vectorwithfusion::as_vector和 MPL 适配器。我已经有一个助手可以将 a 转换fusion::vector为std::tuple。

于 2011-07-09T04:06:43.273 回答