Question

我已经有一个给出tuple的模板，而index_sequence可以创建一个新的tuple，其中包含由该序列索引的元素。此功能模板称为project

我想如果我能以某种方式计算两个index_sequence的差异，那么我就完成了：我将project和make_index_sequence<original_tuple_t>的差异传递给GsonBuilder gsonBuilder = new GsonBuilder(); gsonBuilder = gsonBuilder.setPrettyPrinting(); //Sets pretty formatting Gson gson = gsonBuilder.create(); //Create Gson reference System.out.println(gson.toJson(object)); //prints in pretty format分类和独特。

这是一个好方法吗？你知道如何实现这个吗？

Answer 1

这是一种可能的实施方式。

首先，我们将从一个简单的特征开始，将std::size_t添加到std::index_sequence的前面：

template <std::size_t First, typename Seq>
struct sequence_cat;

template <std::size_t First, std::size_t... Seq>
struct sequence_cat <First, std::index_sequence<Seq...>> {
    using type = std::index_sequence<First, Seq...>;   
};

//helper
template <std::size_t First, typename Seq>
using sequence_cat_t = typename sequence_cat<First, Seq>::type;

现在我们将定义一个名为not_in_sequence的特征，该特征将递归检查两个std::index_sequence的前端并过滤掉第一个中出现的任何内容：

//empty first sequence
template <class First, class Second>
struct not_in_sequence {
    using type = std::index_sequence<>; 
};

//helper
template <class First, class Second>
using not_in_sequence_t = typename not_in_sequence<First, Second>::type;

//filter and recurse
template <std::size_t... First, std::size_t... Second, 
          std::size_t FirstHead, std::size_t SecondHead>
struct not_in_sequence <std::index_sequence<FirstHead, First...>,
                        std::index_sequence<SecondHead, Second...>> {
    using seq1 = std::index_sequence<First...>;
    using seq2 = std::index_sequence<Second...>;

    using type = 
        std::conditional_t<
            (FirstHead == SecondHead),
            not_in_sequence_t<seq1, seq2>,
            sequence_cat_t<
                FirstHead, 
                not_in_sequence_t<seq1, sequence_cat_t<SecondHead, seq2>>
            >
         >;   
}; 
//empty second sequence
template <std::size_t... First, std::size_t FirstHead>
struct not_in_sequence <std::index_sequence<FirstHead, First...>,
                        std::index_sequence<>> {
    using type = std::index_sequence<FirstHead, First...>;   
};

你可以这样使用not_in_sequence_t：

not_in_sequence_t<std::make_index_sequence<3>, std::index_sequence<0,2>>
//same as std::index_sequence<1>

我可能会遗漏一些边缘情况，但您可以随意编辑。

Live Demo

C ++从索引不在给定index_sequence中的元素构造元组

1 个答案: