std::ranges...erge_result(3)

🦋 std::ranges::merge, std::ranges::merge_result 🦋

okay so welcome to the C++ Standard Libary manual page for std::ranges::merge and std::ranges::merge_result — section (3), bestie. this is the one where two sorted ranges get merged into one sorted range, and honestly? iconic. 💅

🌸 NAME 🌸

• std::ranges::merge,std::ranges::merge_result — std::ranges::merge,std::ranges::merge_result

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✨ Synopsis ✨

Defined in header <algorithm>

💖 Call signature 💖

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
>
(1) (since C++20)
requires std::mergeable<I1, I2, O, Comp, Proj1, Proj2>
constexpr merge_result<I1, I2, O>
    merge( I1 first1, S1 last1, I2 first2, S2 last2, O result,
Comp comp = {},

           Proj1 proj1 = {}, Proj2 proj2 = {} );
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>,
(2) (since C++20)
                            O, Comp, Proj1, Proj2>
constexpr merge_result<ranges::borrowed_iterator_t<R1>,
                       ranges::borrowed_iterator_t<R2>, O>
    merge( R1&& r1, R2&& r2, O result, Comp comp = {},

           Proj1 proj1 = {}, Proj2 proj2 = {} );

🎀 Helper types 🎀

template< class I1, class I2, class O >
(3) (since C++20)
using merge_result = ranges::in_in_out_result<I1, I2, O>;

Merges two sorted ranges [[first1, last1) and [first2, last2) into one sorted range beginning at result.

A sequence is said to be sorted with respect to the comparator comp if for any iterator it pointing to the sequence and any non-negative integer n such that it + n is a valid iterator pointing to an element of the sequence, std::invoke(comp, std::invoke(proj2, *(it + n)), std::invoke(proj1, *it))) evaluates to false.

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.

lowkey important, bestie: the behavior is undefined if the destination range overlaps either of the input ranges (the input ranges may overlap each other). like, no cap, don’t do that. 💀

This merge function is stable, which means that for equivalent elements in the original two ranges, the elements from the first range (preserving their original order) precede the elements from the second range (preserving their original order).

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.

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💜 Parameters 💜

• first1, last1 — the first input sorted range.
• first2, last2 — 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 💖

{last1, last2, result_last}, where result_last is the end of the constructed range.

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🔥 Complexity 🔥

At most N − 1 comparisons and applications of each projection, where N = ranges::distance(first1, last1) + ranges::distance(first2, last12).

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🧚 Notes 🧚

This algorithm performs a similar task as ranges::set_union 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 max(n, m) ones only. So ranges::merge outputs exactly N values and ranges::set_union may produce fewer.

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🌈 Possible implementation 🌈

struct merge_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::merge_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(proj2, *first2),
std::invoke(proj1, *first1)))
                *result = *first2, ++first2;
            else
                *result = *first1, ++first1;
        }
        auto ret1{ranges::copy(std::move(first1), std::move(last1),
std::move(result))};
        auto ret2{ranges::copy(std::move(first2), std::move(last2),
std::move(ret1.out))};
        return {std::move(ret1.in), std::move(ret2.in),
std::move(ret2.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::merge_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 merge_fn merge {};

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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 << "} +\n{ ";
    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::merge(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::merge(in1, in2, std::back_inserter(out));
    print(in1, in2, out.cbegin(), out.cend());
}

🌟 Output 🌟

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

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

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👑 See also 👑

• ranges::inplace_merge — merges two ordered ranges in-place (C++20) (niebloid)
• ranges::is_sorted — checks whether a range is sorted into ascending order (C++20) (niebloid)
• ranges::set_union — computes the union of two sets (C++20) (niebloid)
• ranges::sort — sorts a range into ascending order (C++20) (niebloid)
• ranges::stable_sort — sorts a range of elements while preserving order between equal elements (C++20) (niebloid)
• merge — merges two sorted ranges (function template)

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okay bestie that’s the whole page 💖 source: http://cppreference.com • version/date: 2024.06.10 • page: std::ranges...erge_result(3) ✨ bye, slay, and remember: sorted inputs only 🫶