Question

我想允许使用我正在编写的类来指定类型列表作为模板参数以及这些类型的分配器列表，其方式是类型位于奇数位置且分配器位于偶数位置：

template<typename... T>
class MyClass {
  // Stuff inside
}

int main() {
  MyClass<SomeType1, AllocatorOfSomeType1> c1;
  MyClass<SomeType1, AllocatorOfSomeType1, 
          SomeType2, AllocatorOfSomeType2> c2;
  MyClass<SomeType1, AllocatorOfSomeType1, 
          SomeType2, AllocatorOfSomeType2,
          SomeType3, AllocatorOfSomeType3> c3;
  // And so on....
}

在内部，有一个存储类型的向量元组是有意义的：

std::tuple<std::vector<EveryOddTypeInParameterPack>...> m_storage_;

以及用于使用的分配器元组：

std::tuple<std::vector<EveryEvenTypeInParameterPack>...> m_storage_;

如何在代码中实际声明这些元组？理论上我需要以某种方式选择参数包中的每个奇/偶类型 - 这可能吗？

Answer 1

虽然代码有点冗长，但我认为机制没有不必要的特点。
如果我正确理解了这个问题，可能以下代码将符合目的：

// push front for tuple
template< class, class > struct PFT;

template< class A, class... T > struct PFT< A, tuple< T... > > {
  typedef tuple< A, T... > type;
};

// for even
template< class... > struct even_tuple;

template< class A, class B > struct even_tuple< A, B > {
  typedef tuple< A > type;
};
template< class A, class B, class... T > struct even_tuple< A, B, T... > {
  typedef typename PFT< A, typename even_tuple< T... >::type >::type type;
};
// As for odd elements, in the same way as even(please see the test on ideone)

// objective type
template< class > struct storage_type;

template< class... T > struct storage_type< tuple< T... > > {
  typedef tuple< vector< T >... > type;
};

template< class... T >
struct MyClass {
  typename storage_type< typename even_tuple< T... >::type >::type
    m_storage_even_;
  typename storage_type< typename  odd_tuple< T... >::type >::type
    m_storage_odd_;
};

以下是ideone的测试。

Answer 2

也许是这样的：

#include <tuple>

// Example receptacle    
template <typename ...Args> struct MyContainer;

// Tuple concatenator
template<typename PackR, typename PackL> struct cat;
template<typename ...R, typename ...L>
struct cat<std::tuple<R...>, std::tuple<L...>>
{
  typedef std::tuple<R..., L...> type;
};

// Even/Odd extractors
template <typename ...Args> struct GetEven;
template <typename ...Args> struct GetOdd;

template <typename E1, typename O1, typename ...Args>
struct GetEven<E1, O1, Args...>
{
  typedef typename cat<std::tuple<E1>, typename GetEven<Args...>::value>::type value;
};
template <typename E1, typename O1>
struct GetEven<E1, O1>
{
  typedef std::tuple<E1> value;
};

template <typename E1, typename O1, typename ...Args>
struct GetOdd<E1, O1, Args...>
{
  typedef typename cat<std::tuple<O1>, typename GetEven<Args...>::value>::type value;
};
template <typename E1, typename O1>
struct GetOdd<E1, O1>
{
  typedef std::tuple<O1> value;
};

// Tuple-to-Receptacle mover
template <typename Pack, template <typename ...T> class Receiver> struct Unpack;
template <typename ...Args, template <typename ...T> class Receiver>
struct Unpack<std::tuple<Args...>, Receiver>
{
  typedef Receiver<Args...> type;
};

// Example consumer
template <typename ...Args>
struct Foo
{
  typedef typename Unpack<typename GetEven<Args...>::value, MyContainer>::type EvenVector;
  typedef typename Unpack<typename GetOdd<Args...>::value, MyContainer>::type OddVector;

  EvenVector x;
  OddVector y;
};

您仍然需要定义MyContainer类来对可变参数执行一些有用的操作，例如：实现你的向量元组...（为什么不是元组的向量，但是？）

对于元组技巧的归属brunocodutra。

Answer 3

这只是一次尝试

template<typename... T> class Myclass;

template<typename T1, typename allocT1>
class MyClass <T1, allocT1> {
  std::pair<T1, allocT1> myFirstArglist;
//and you have to do a check that allocT1::value_type is same as T1 or not
//or may be alloT1 is an allocator type or not(i'm thinking concepts, may be)
//this idea is inspired from Chris's comment
};

template<typename T1, typename allocT1, typename... T>
class Myclass<T1, allocT1, T...> {
std::pair<T1, allocT1> myFirstArglist;
Myclass<T>; //something like this
};

template<>
class Myclass<> {
//probably you would like some error message here
//when there are no types and containers
};

可能我不够清楚，你可能想读 http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2006/n2080.pdf

还有一个与分配器类型设计相关的好帖子......你想看看： C++ Design Pattern for allocator type arguments

Answer 4

我知道您的问题最初被标记为“ c ++ 11”，但我认为值得一提的是，在C ++ 14中您可以访问make_index_sequence，这使得整个过程变得非常简单。为了过滤元组，我将从以下概述开始：https://quuxplusone.github.io/blog/2018/07/23/metafilter/

然后我们得到这样的内容（Godbolt）：

template<bool> struct zero_or_one {
    template<class E> using type = std::tuple<E>;
};

template<> struct zero_or_one<false> {
    template<class E> using type = std::tuple<>;
};

template<class Tuple, class = std::make_index_sequence<std::tuple_size<Tuple>::value>>
struct just_evens;

template<class... Es, size_t... Is>
struct just_evens<std::tuple<Es...>, std::index_sequence<Is...>> {
    using type = decltype(std::tuple_cat(
        std::declval<typename zero_or_one<Is % 2 == 0>::template type<Es>>()...
    ));
};

要获取just_odds，您需要将条件从Is % 2 == 0切换到Is % 2 != 0。

用法示例：

static_assert(std::is_same<
    just_evens<std::tuple<char, short, int, long, double>>::type,
    std::tuple<char, int, double>
>::value, "");

选择模板参数包中的每个偶数（或奇数）参数

4 个答案: