std::ranges...nion_result(3)

✨ NAME ✨

okay so bestie, this page is literally about std::ranges::set_union and std::ranges::set_union_result, and the title up top is the iconic little std::ranges...nion_result(3) from the C++ Standard Libary (yes that spelling is a mood, we’re keeping it). the NAME line in the original is std::ranges::set_union,std::ranges::set_union_result - std::ranges::set_union,std::ranges::set_union_result, which is giving “I said what I said” energy. 💅

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✨ Synopsis ✨

Defined in header <algorithm>.

Call signature time, bestie:

(1) (since C++20)

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_union_result<I1, I2, O>
    set_union( I1 first1, S1 last1, I2 first2, S2 last2,
               O result, Comp comp = {},
               Proj1 proj1 = {}, Proj2 proj2 = {} );

(2) (since C++20)

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_union_result<ranges::borrowed_iterator_t<R1>,
                           ranges::borrowed_iterator_t<R2>, O>
    set_union( R1&& r1, R2&& r2, O result, Comp comp = {},
               Proj1 proj1 = {}, Proj2 proj2 = {} );

Helper types:

(3) (since C++20)

template< class I1, class I2, class O >
    using set_union_result = ranges::in_in_out_result<I1, I2, O>;

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✨ Description ✨

so what does it do? it constructs a sorted union beginning at result consisting of the set of elements present in one or both sorted input ranges [first1, last1) and [first2, last2). ✨

and now the multiplicity tea ☕: if some element is found m times in [first1, last1) and n times in [first2, last2), then all m elements will be copied from [first1, last1) to result, preserving order, and then exactly max(n - m, 0) elements will be copied from [first2, last2) to result, also preserving order. periodt.

🚨 the behavior is undefined if: - the input ranges are not sorted with respect to comp and proj1 or proj2, respectively, or - the resulting range overlaps with either of the input ranges.

okay so like: (1) elements are compared using the given binary comparison function comp.

(2) same as (1), but uses r1 as the first range and r2 as the second range, as if using ranges::begin(r1) as first1, ranges::end(r1) as last1, ranges::begin(r2) as first2, and ranges::end(r2) as last2.

the function-like entities described on this page are niebloids, that is:

• explicit template argument lists cannot be specified when calling any of them.
• none of them are visible to argument-dependent lookup.
• when any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

in practice, they may be implemented as function objects, or with special compiler extensions. no cap. 🫶

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✨ Parameters ✨

• first1, last1 — iterator-sentinel pair denoting the first input sorted range
• first2, last2 — iterator-sentinel pair denoting the second input sorted range
• r1 — the first input sorted range
• r2 — the second input sorted range
• result — the beginning of the output range
• comp — comparison to apply to the projected elements
• proj1 — projection to apply to the elements in the first range
• proj2 — projection to apply to the elements in the second range

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✨ Return value ✨

it returns {last1, last2, result_last}, where result_last is the end of the constructed range. slay. 💖

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✨ Complexity ✨

at most 2·(N1+N2)-1 comparisons and applications of each projection, where N1 and N2 are ranges::distance(first1, last1) and ranges::distance(first2, last2), respectively. math girlie is mathing. 🔥

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✨ Notes ✨

this algorithm performs a similar task as ranges::merge does. both consume two sorted input ranges and produce a sorted output with elements from both inputs. the difference between these two algorithms is with handling values from both input ranges which compare equivalent (see notes on LessThanComparable). if any equivalent values appeared n times in the first range and m times in the second, ranges::merge would output all n+m occurrences whereas ranges::set_union would output std::max(n, m) ones only. so ranges::merge outputs exactly (N1+N2) values and ranges::set_union may produce less. iykyk. 🌸

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✨ Possible implementation ✨

struct set_union_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_union_result<I1, I2, O>
        operator()(I1 first1, S1 last1, I2 first2, S2 last2,
                   O result, Comp comp = {},
                   Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        for (; !(first1 == last1 or first2 == last2); ++result)
        {
            if (std::invoke(comp, std::invoke(proj1, *first1),
                                  std::invoke(proj2, *first2)))
            {
                *result = *first1;
                ++first1;
            }
            else if (std::invoke(comp, std::invoke(proj2, *first2),
                                       std::invoke(proj1, *first1)))
            {
                *result = *first2;
                ++first2;
            }
            else
            {
                *result = *first1;
                ++first1;
                ++first2;
            }
        }
        auto res1 = ranges::copy(std::move(first1), std::move(last1),
       std::move(result));
        auto res2 = ranges::copy(std::move(first2), std::move(last2),
       std::move(res1.out));
        return {std::move(res1.in), std::move(res2.in),
       std::move(res2.out)};
    }

    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_union_result<ranges::borrowed_iterator_t<R1>,
                                       ranges::borrowed_iterator_t<R2>,
       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_union_fn set_union {};

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✨ Example ✨

// Run this code

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

void print(const auto& in1, const auto& in2, auto first, auto last)
{
    std::cout << "{ ";
    for (const auto& e : in1)
        std::cout << e << ' ';
    std::cout << "} ∪ { ";
    for (const auto& e : in2)
        std::cout << e << ' ';
    std::cout << "} =\n{ ";
    while (!(first == last))
        std::cout << *first++ << ' ';
    std::cout << "}\n\n";
}

int main()
{
    std::vector<int> in1, in2, out;

    in1 = {1, 2, 3, 4, 5};
    in2 = {      3, 4, 5, 6, 7};
    out.resize(in1.size() + in2.size());
    const auto ret = std::ranges::set_union(in1, in2, out.begin());
    print(in1, in2, out.begin(), ret.out);

    in1 = {1, 2, 3, 4, 5, 5, 5};
    in2 = {      3, 4, 5, 6, 7};
    out.clear();
    out.reserve(in1.size() + in2.size());
    std::ranges::set_union(in1, in2, std::back_inserter(out));
    print(in1, in2, out.cbegin(), out.cend());
}

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✨ Output ✨

{ 1 2 3 4 5 } ∪ { 3 4 5 6 7 } =
{ 1 2 3 4 5 6 7 }

{ 1 2 3 4 5 5 5 } ∪ { 3 4 5 6 7 } =
{ 1 2 3 4 5 5 5 6 7 }

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✨ See also ✨

• ranges::set_difference (C++20) — computes the difference between two sets (niebloid)
• ranges::set_intersection (C++20) — computes the intersection of two sets (niebloid)
• ranges::set_symmetric_difference (C++20) — computes the symmetric difference between two sets (niebloid)
• ranges::merge (C++20) — merges two sorted ranges (niebloid)
• ranges::includes (C++20) — returns true if one sequence is a subsequence of another (niebloid)
• set_union — computes the union of two sets (function template)

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and that’s the whole page, bestie 💜 source is http://cppreference.com, version 2024.06.10, and the footer title is std::ranges...nion_result(3) — go forth and union your sorted ranges, but only if they’re actually sorted, fr fr. 🦋