18

我最近遇到了这个难题,终于能够找出一个骇人听闻的答案(使用索引数组),并想分享它(答案如下)。我确信有使用模板递归的答案和使用boost; 如果您有兴趣,请分享其他方法来做到这一点。我认为将这些全部放在一个地方可能会使其他人受益,并且对于学习一些很酷的 C++11 模板元编程技巧很有用。

问题: 给定两个长度相等的元组:

auto tup1 = std::make_tuple(1, 'b', -10);
auto tup2 = std::make_tuple(2.5, 2, std::string("even strings?!"));

您如何创建一个将两个元组“压缩”成对的异构元组的函数?

std::tuple<
    std::pair<int, double>,
    std::pair<char, int>,
    std::pair<int, std::string> > result =
    tuple_zip( tup1, tup2 );

在哪里

std::get<0>(result) == std::make_pair(1, 2.5);
std::get<1>(result) == std::make_pair('b', 2);
std::get<2>(result) == std::make_pair(-10, std::string("even strings?!"));
4

2 回答 2

18

首先,快速浏览索引数组:

template<std::size_t ...S>
struct seq { };

// And now an example of how index arrays are used to print a tuple:
template <typename ...T, std::size_t ...S>
void print_helper(std::tuple<T...> tup, seq<S...> s) {
  // this trick is exceptionally useful:
  // ((std::cout << std::get<S>(tup) << " "), 0) executes the cout
  // and returns 0.
  // { 0... } expands (because the expression has an S in it),
  // returning an array of length sizeof...(S) full of zeros.
  // The array isn't used, but it's a great hack to do one operation
  // for each std::size_t in S.
  int garbage[] = { ((std::cout << std::get<S>(tup) << " "), 0)... };
  std::cout << std::endl;
}

现在使用我们的 print_helper 函数:

int main() {
  print_helper(std::make_tuple(10, 0.66, 'h'), seq<0,1,2>() );
  return 0;
}

不过,打字seq<0,1,2>可能会有点痛苦。所以我们可以使用模板递归来创建一个类来生成seqs,这样gens<3>::type就和seq<0,1,2>

template<std::size_t N, std::size_t ...S>
struct gens : gens<N-1, N-1, S...> { };

template<std::size_t ...S>
struct gens<0, S...> {
  typedef seq<S...> type;
};

int main() {
  print_helper(std::make_tuple(10, 0.66, 'h'), gens<3>::type() );
  return 0;
}

由于Ningens<N>::type始终是元组中元素的数量,因此您可以换print_helper行以使其更容易:

template <typename ...T>
void print(std::tuple<T...> tup) {
  print_helper(tup, typename gens<sizeof...(T)>::type() );
}

int main() {
  print(std::make_tuple(10, 0.66, 'h'));
  return 0;
}

请注意,模板参数可以自动推导出(输入所有这些会很痛苦,不是吗?)。

现在,tuple_zip功能:

和以前一样,从辅助函数开始:

template <template <typename ...> class Tup1,
    template <typename ...> class Tup2,
    typename ...A, typename ...B,
    std::size_t ...S>
auto tuple_zip_helper(Tup1<A...> t1, Tup2<B...> t2, seq<S...> s) ->
decltype(std::make_tuple(std::make_pair(std::get<S>(t1),std::get<S>(t2))...)) {
  return std::make_tuple( std::make_pair( std::get<S>(t1), std::get<S>(t2) )...);
}

代码有点棘手,尤其是尾随返回类型(返回类型声明为auto->在参数定义后提供)。这让我们避免了定义返回类型的问题,只需声明它返回函数体中使用的表达式(如果xyints,delctype(x+y)在编译时解析为int)。

现在将它包装在一个提供适当seq<0, 1...N>using的函数中gens<N>::type

template <template <typename ...> class Tup1,
  template <typename ...> class Tup2,
  typename ...A, typename ...B>
auto tuple_zip(Tup1<A...> t1, Tup2<B...> t2) ->
decltype(tuple_zip_helper(t1, t2, typename gens<sizeof...(A)>::type() )) {
  static_assert(sizeof...(A) == sizeof...(B), "The tuple sizes must be the same");
  return tuple_zip_helper( t1, t2, typename gens<sizeof...(A)>::type() );
}

现在您可以按照问题中的说明使用它:

int main() {
  auto tup1 = std::make_tuple(1, 'b', -10);
  auto tup2 = std::make_tuple(2.5, 2, std::string("even strings?!"));
  std::tuple<
    std::pair<int, double>,
    std::pair<char, int>,
    std::pair<int, std::string> > x = tuple_zip( tup1, tup2 );

  // this is also equivalent:
  //  auto x = tuple_zip( tup1, tup2 );

  return 0;
}

最后,如果你提供一个<<操作符,std::pair你可以使用我们上面定义的 print 函数来打印压缩后的结果:

template <typename A, typename B>
std::ostream & operator << (std::ostream & os, const std::pair<A, B> & pair) {
  os << "pair("<< pair.first << "," << pair.second << ")";
  return os;
}

int main() {
  auto tup1 = std::make_tuple(1, 'b', -10);
  auto tup2 = std::make_tuple(2.5, 2, std::string("even strings?!"));
  auto x = tuple_zip( tup1, tup2 );

  std::cout << "zipping: ";
  print(tup1);
  std::cout << "with   : ";
  print(tup2);

  std::cout << "yields : ";
  print(x);

  return 0;
}

输出是:

zipping: 1 b 10
with : 2.5 2 even strings?!
产生:pair(1,2.5) pair(b,2) pair(10,even strings?!)

Like std::array,std::tuple是在编译时定义的,因此它可用于生成更可优化的代码(与 和 之类的容器相比,在编译时已知更多信息std::vectorstd::list。因此,即使有时需要做一些工作,有时您也可以使用它来编写快速而聪明的代码。快乐黑客!


编辑:

根据要求,允许不同大小的元组和空指针填充:

template <typename T, std::size_t N, std::size_t ...S>
auto array_to_tuple_helper(const std::array<T, N> & arr, seq<S...> s) -> decltype(std::make_tuple(arr[S]...)) {
  return std::make_tuple(arr[S]...);
}

template <typename T, std::size_t N>
auto array_to_tuple(const std::array<T, N> & arr) -> decltype( array_to_tuple_helper(arr, typename gens<N>::type()) ) {
  return array_to_tuple_helper(arr, typename gens<N>::type());
}

template <std::size_t N, template <typename ...> class Tup, typename ...A>
auto pad(Tup<A...> tup) -> decltype(tuple_cat(tup, array_to_tuple(std::array<std::nullptr_t, N>()) )) {
  return tuple_cat(tup, array_to_tuple(std::array<std::nullptr_t, N>()) );
}

#define EXTENSION_TO_FIRST(first,second) ((first)>(second) ? (first)-(second) : 0)

template <template <typename ...> class Tup1, template <typename ...> class Tup2, typename ...A, typename ...B>
auto pad_first(Tup1<A...> t1, Tup2<B...> t2) -> decltype( pad<EXTENSION_TO_FIRST(sizeof...(B), sizeof...(A)), Tup1, A...>(t1) ) {
  return pad<EXTENSION_TO_FIRST(sizeof...(B), sizeof...(A)), Tup1, A...>(t1);
}

template <template <typename ...> class Tup1, template <typename ...> class Tup2, typename ...A, typename ...B>
auto diff_size_tuple_zip(Tup1<A...> t1, Tup2<B...> t2) ->
  decltype( tuple_zip( pad_first(t1, t2), pad_first(t2, t1) ) ) {
  return tuple_zip( pad_first(t1, t2), pad_first(t2, t1) );
}

顺便说一句,你现在需要这个来使用我们方便的print功能:

std::ostream & operator << (std::ostream & os, std::nullptr_t) {
  os << "null_ptr";
  return os;
}
于 2012-07-04T03:32:17.350 回答
2

对于任意数量的元组来说,这样做并不是非常困难。

一种方法是创建一个函数,将 N 个元组中特定索引处的所有元素收集到一个新元组中。然后有另一个函数将这些元组收集到原始元组中每个索引的新元组中。

所有这些都可以通过使用参数包扩展表达式来相对简单地完成,而无需任何递归函数。

#include <cstddef>
#include <tuple>

namespace detail {
    // Describe the type of a tuple with element I from each input tuple.
    // Needed to preserve the exact types from the input tuples.
    template<std::size_t I, typename... Tuples>
    using zip_tuple_at_index_t = std::tuple<std::tuple_element_t<I, std::decay_t<Tuples>>...>;

    // Collect all elements at index I from all input tuples as a new tuple.
    template<std::size_t I, typename... Tuples>
    zip_tuple_at_index_t<I, Tuples...> zip_tuple_at_index(Tuples && ...tuples) {
        return {std::get<I>(std::forward<Tuples>(tuples))...};
    }

    // Create a tuple with the result of zip_tuple_at_index for each index.
    // The explicit return type prevents flattening into a single tuple
    // when sizeof...(Tuples) == 1 or sizeof...(I) == 1 .
    template<typename... Tuples, std::size_t... I>
    std::tuple<zip_tuple_at_index_t<I, Tuples...>...> tuple_zip_impl(Tuples && ...tuples, std::index_sequence<I...>) {
        return {zip_tuple_at_index<I>(std::forward<Tuples>(tuples)...)...};
    }

}

// Zip a number of tuples together into a tuple of tuples.
// Take the first tuple separately so we can easily get its size.
template<typename Head, typename... Tail>
auto tuple_zip(Head && head, Tail && ...tail) {
    constexpr std::size_t size = std::tuple_size_v<std::decay_t<Head>>;

    static_assert(
        ((std::tuple_size_v<std::decay_t<Tail>> == size) && ...),
        "Tuple size mismatch, can not zip."
    );

    return detail::tuple_zip_impl<Head, Tail...>(
        std::forward<Head>(head),
        std::forward<Tail>(tail)...,
        std::make_index_sequence<size>()
    );
}

在此处查看实际操作:https ://wandbox.org/permlink/EQhvLPyRfDrtjDMw

我使用了一些 C++14/17 特性,但没有什么必要的。最难替换的部分是用于检查元组大小的折叠表达式。那可能必须成为递归检查。

于 2017-11-05T21:51:55.330 回答