# std::ranges::set_difference, std::ranges::set_difference_result

< cpp‎ | algorithm‎ | ranges

定义于头文件 `<algorithm>` |  |
调用签名 |  |
template< std::input_iterator I1, std::sentinel_for<I1> S1, std::input_iterator I2, std::sentinel_for<I2> S2, std::weakly_incrementable O, class Comp = ranges::less, class Proj1 = std::identity, class Proj2 = std::identity > requires std::mergeable<I1, I2, O, Comp, Proj1, Proj2> constexpr set_difference_result<I1, O> set_difference( I1 first1, S1 last1, I2 first2, S2 last2, O result, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ); | (1) | (C++20 起)
template< ranges::input_range R1, ranges::input_range R2, std::weakly_incrementable O, class Comp = ranges::less, class Proj1 = std::identity, class Proj2 = std::identity > requires std::mergeable<ranges::iterator_t<R1>, ranges::iterator_t<R2>, O, Comp, Proj1, Proj2> constexpr set_difference_result<ranges::borrowed_iterator_t<R1>, O> set_difference( R1&& r1, R2&& r2, O result, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ); | (2) | (C++20 起)
辅助类型 |  |
template<class I, class O> using set_difference_result = ranges::in_out_result<I, O>; | (3) | (C++20 起)

从已排序的输入范围 `[first1, last1)` 赋值在已排序的输入范围 `[first2, last2)` 中找不到的元素到始于 `result` 的输出范围。

若

- 输入范围未分别相对于 `comp` 与 `proj1` 或 `proj2` 排序，或
- 结果范围与任一输入范围重叠，

则行为未定义。

1) 用给定的二元比较函数 `comp` 比较元素。
2) 同 (1) ，但以 `r1` 为第一范围，并以 `r2` 为第二范围，如同以 ranges::begin(r1) 为 `first1` ，以 ranges::end(r1) 为 `last1` ，以 ranges::begin(r2) 为 `first2` ，并以 ranges::end(r2) 为 `last2` 。

此页面上描述的仿函数实体是 niebloid ，即：

- 调用它们中的任何一者时不可指定显式模板实参。
- 它们无一对实参依赖查找可见。
- 在通常无限定查找寻找函数调用运算符左侧的名称找到它们之一时，它禁止实参依赖查找。

实际上，它们能以函数对象，或以某些特殊编译器扩展实现。

### 参数

first1, last1 | - | 第一已排序输入范围
first2, last2 | - | 第二已排序输入范围
r1 | - | 第一已排序输入范围
r2 | - | 第二已排序输入范围
result | - | 目标范围的起始
comp | - | 应用到投影后元素的比较器
proj1 | - | 应用到第一范围元素的投影
proj2 | - | 应用到第二范围元素的投影

### 返回值

{last1, result_last} ，其中 result_last 为被构造范围末尾。

### 复杂度

至多比较和应用每个投影 \(\scriptsize 2\cdot(N_1+N_2)-1\)2·(N
1+N
2)-1 次，其中 \(\scriptsize N_1\)N
1 与 \(\scriptsize N_2\)N
2 分别为 ranges::distance(first1, last1) 与 ranges::distance(first2, last2) 。

### 可能的实现

    struct set_difference_fn {
        template< std::input_iterator I1, std::sentinel_for<I1> S1,
                  std::input_iterator I2, std::sentinel_for<I2> S2,
                  std::weakly_incrementable O, class Comp = ranges::less,
                  class Proj1 = std::identity, class Proj2 = std::identity >
        requires std::mergeable<I1, I2, O, Comp, Proj1, Proj2>
        constexpr ranges::set_difference_result<I1, O>
        operator()( I1 first1, S1 last1, I2 first2, S2 last2,
                    O result, Comp comp = {},
                    Proj1 proj1 = {}, Proj2 proj2 = {} ) const {
            while (!(first1 == last1 or first2 == last2)) {
                if (std::invoke(comp, std::invoke(proj1, *first1),
                                      std::invoke(proj2, *first2))) {
                    *result = *first1; ++first1; ++result;
                }
                else if (std::invoke(comp, std::invoke(proj2, *first2),
                                     std::invoke(proj1, *first1))) {
                    ++first2;
                }
                else {
                    ++first1; ++first2;
                }
            }
            return ranges::copy(std::move(first1), std::move(last1), std::move(result));
        }

        template< ranges::input_range R1, ranges::input_range R2,
                  std::weakly_incrementable O, class Comp = ranges::less,
                  class Proj1 = std::identity, class Proj2 = std::identity >
        requires std::mergeable<ranges::iterator_t<R1>, ranges::iterator_t<R2>,
                 O, Comp, Proj1, Proj2>
        constexpr ranges::set_difference_result<
                  ranges::borrowed_iterator_t<R1>, O>
        operator()( R1&& r1, R2&& r2, O result, Comp comp = {},
                    Proj1 proj1 = {}, Proj2 proj2 = {} ) const {
            return (*this)(ranges::begin(r1), ranges::end(r1),
                           ranges::begin(r2), ranges::end(r2),
                           std::move(result), std::move(comp),
                           std::move(proj1), std::move(proj2));
        }
    };

    inline constexpr set_difference_fn set_difference{};

### 示例

    #include <algorithm>
    #include <cassert>
    #include <iostream>
    #include <iterator>
    #include <string_view>
    #include <vector>

    auto print = [](const auto& v, std::string_view end = "") {
        for (std::cout << "{ "; auto i : v) std::cout << i << ' ';
        std::cout << "} " << end;
    };

    struct Order // 拥有非常有趣的数据的类
    {
        int order_id;

        friend std::ostream& operator<<(std::ostream& os, const Order& ord) {
            return os << "{" << ord.order_id << "},";
        }
    };

    int main()
    {
        const auto v1 = {1, 2, 5, 5, 5, 9};
        const auto v2 = {2, 5, 7};
        std::vector<int> diff;

        std::ranges::set_difference(v1, v2, std::back_inserter(diff));
        print(v1, "∖ ");
        print(v2, "= ");
        print(diff, "\n");

        // 我们想知道哪些顺序在旧和新状态之间“改变了”：
        const std::vector<Order> old_orders { {1}, {2}, {5}, {9}, };
        const std::vector<Order> new_orders { {2}, {5}, {7}, };
        std::vector<Order> cut_orders(old_orders.size() + new_orders.size());

        auto [old_orders_end, cut_orders_last] =
            std::ranges::set_difference(old_orders, new_orders,
                                        cut_orders.begin(), {},
                                        &Order::order_id, &Order::order_id);
        assert(old_orders_end == old_orders.end());

        std::cout << "old orders = "; print(old_orders, "\n");
        std::cout << "new orders = "; print(new_orders, "\n");
        std::cout << "cut orders = "; print(cut_orders, "\n");
        cut_orders.erase(cut_orders_last, end(cut_orders));
        std::cout << "cut orders = "; print(cut_orders, "\n");
    }

输出：

    { 1 2 5 5 5 9 } ∖ { 2 5 7 } = { 1 5 5 9 }
    old orders = { {1}, {2}, {5}, {9}, }
    new orders = { {2}, {5}, {7}, }
    cut orders = { {1}, {9}, {0}, {0}, {0}, {0}, {0}, }
    cut orders = { {1}, {9}, }

### 参阅

ranges::set_union (C++20) | 计算两个集合的并集 (niebloid)
ranges::set_intersection (C++20) | 计算两个集合的交集 (niebloid)
ranges::set_symmetric_difference (C++20) | 计算两个集合的对称差 (niebloid)
ranges::includes (C++20) | 若一个序列是另一个的子列则返回 `true` (niebloid)
set_difference | 计算两个集合的差集 (函数模板)