std::merge(3)

💅 std::merge(3) — C++ Standard Library 💅

okay so bestie, this is the std::merge(3) page from the C++ Standard Library manual. It’s giving “merge two sorted ranges into one sorted range” energy, and honestly? iconic algorithm behavior. Let’s get into it ✨

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🌸 NAME 🌸

std::merge — std::merge

Not me explaining the NAME section like it’s deep, but yes: that’s literally the whole name listing 🫶

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🦋 Synopsis 🦋

Defined in header <algorithm>:

(1) (constexpr since C++20)

template< class InputIt1, class InputIt2, class OutputIt >
OutputIt merge( InputIt1 first1, InputIt1 last1,
                InputIt2 first2, InputIt2 last2,
                OutputIt d_first );

(2) (since C++17)

template< class ExecutionPolicy,
          class ForwardIt1, class ForwardIt2, class ForwardIt3 >
ForwardIt3 merge( ExecutionPolicy&& policy,
                  ForwardIt1 first1, ForwardIt1 last1,
                  ForwardIt2 first2, ForwardIt2 last2,
                  ForwardIt3 d_first );

(3) (constexpr since C++20)

template< class InputIt1, class InputIt2,
          class OutputIt, class Compare >
OutputIt merge( InputIt1 first1, InputIt1 last1,
                InputIt2 first2, InputIt2 last2,
                OutputIt d_first, Compare comp );

(4) (since C++17)

template< class ExecutionPolicy,
          class ForwardIt1, class ForwardIt2,
          class ForwardIt3, class Compare >
ForwardIt3 merge( ExecutionPolicy&& policy,
                  ForwardIt1 first1, ForwardIt1 last1,
                  ForwardIt2 first2, ForwardIt2 last2,
                  ForwardIt3 d_first, Compare comp );

Merges two sorted ranges [first1, last1) and [first2, last2) into one sorted range beginning at d_first. Like two sorted besties linking arms and becoming one sorted supergroup 💖

• 1) If [first1, last1) or [first2, last2) is not sorted with respect to operator< (until C++20) / std::less{} (since C++20), the behavior is undefined.
• 3) If [first1, last1) or [first2, last2) is not sorted with respect to comp, the behavior is undefined.
• 2,4) Same as (1,3), but executed according to policy. These overloads participate in overload resolution only if std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> is true. (until C++20) std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>> is true. (since C++20)

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). Slay for order preservation 💅

If the output range overlaps with [first1, last1) or [first2, last2), the behavior is undefined. No cap, do not overlap 🙈

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🌸 Parameters 🌸

• first1, last1 — the first range of elements to merge.
• first2, last2 — the second range of elements to merge.
• d_first — the beginning of the destination range.
• policy — the execution policy to use. See execution policy for details.
• comp — comparison function object (i.e. an object that satisfies the requirements of Compare) which returns true if the first argument is less than (i.e. is ordered before) the second.

The signature of the comparison function should be equivalent to the following:

bool cmp(const Type1& a, const Type2& b);

While the signature does not need to have const&, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) Type1 and Type2 regardless of value category (thus, Type1& is not allowed, nor is Type1 unless for Type1 a move is equivalent to a copy (since C++11)). The types Type1 and Type2 must be such that objects of types InputIt1 and InputIt2 can be dereferenced and then implicitly converted to both Type1 and Type2.

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✨ Type requirements ✨

• InputIt1, InputIt2 must meet the requirements of LegacyInputIterator.
• ForwardIt1, ForwardIt2, ForwardIt3 must meet the requirements of LegacyForwardIterator.
• OutputIt must meet the requirements of LegacyOutputIterator.
• Compare must meet the requirements of Compare.

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🎀 Return value 🎀

An output iterator to element past the last element copied.

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🔥 Complexity 🔥

Given N_1 as std::distance(first1, last1) and N_2 as std::distance(first2, last2):

• 1) At most N_1+N_2-1 comparisons using operator< (until C++20) / std::less{} (since C++20).
• 2) O(N_1+N_2) comparisons using operator< (until C++20) / std::less{} (since C++20).
• 3) At most N_1+N_2-1 applications of the comparison function comp.
• 4) O(N_1+N_2) applications of the comparison function comp.

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💀 Exceptions 💀

The overloads with a template parameter named ExecutionPolicy report errors as follows:

• If execution of a function invoked as part of the algorithm throws an exception and ExecutionPolicy is one of the standard policies, std::terminate is called. For any other ExecutionPolicy, the behavior is implementation-defined.
• If the algorithm fails to allocate memory, std::bad_alloc is thrown.

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🧚 Possible implementation 🧚

See also the implementations in libstdc++ and libc++.

merge (1)

template<class InputIt1, class InputIt2, class OutputIt>
OutputIt merge(InputIt1 first1, InputIt1 last1,
               InputIt2 first2, InputIt2 last2,
               OutputIt d_first)
{
    for (; first1 != last1; ++d_first)
    {
        if (first2 == last2)
            return std::copy(first1, last1, d_first);

        if (*first2 < *first1)
        {
            *d_first = *first2;
            ++first2;
        }
        else
        {
            *d_first = *first1;
            ++first1;
        }
    }
    return std::copy(first2, last2, d_first);
}

merge (3)

template<class InputIt1, class InputIt2,
         class OutputIt, class Compare>
OutputIt merge(InputIt1 first1, InputIt1 last1,
               InputIt2 first2, InputIt2 last2,
               OutputIt d_first, Compare comp)
{
    for (; first1 != last1; ++d_first)
    {
        if (first2 == last2)
            return std::copy(first1, last1, d_first);

        if (comp(*first2, *first1))
        {
            *d_first = *first2;
            ++first2;
        }
        else
        {
            *d_first = *first1;
            ++first1;
        }
    }
    return std::copy(first2, last2, d_first);
}

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🍓 Notes 🍓

This algorithm performs a similar task as std::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, std::merge would output all n + m occurrences whereas std::set_union would output std::max(n, m) ones only. So std::merge outputs exactly std::distance(first1, last1) + std::distance(first2, last2) values and std::set_union may produce fewer.

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🌈 Example 🌈

Run this code, bestie:

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

auto print = [](const auto rem, const auto& v)
{
    std::cout << rem;
    std::copy(v.begin(), v.end(), std::ostream_iterator<int>(std::cout, " "));
    std::cout << '\n';
};

int main()
{
    // fill the vectors with random numbers
    std::random_device rd;
    std::mt19937 mt(rd());
    std::uniform_int_distribution<> dis(0, 9);

    std::vector<int> v1(10), v2(10);
    std::generate(v1.begin(), v1.end(), std::bind(dis, std::ref(mt)));
    std::generate(v2.begin(), v2.end(), std::bind(dis, std::ref(mt)));

    print("Originally:\nv1: ", v1);
    print("v2: ", v2);

    std::sort(v1.begin(), v1.end());
    std::sort(v2.begin(), v2.end());

    print("After sorting:\nv1: ", v1);
    print("v2: ", v2);

    // merge
    std::vector<int> dst;
    std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(),
               std::back_inserter(dst));

    print("After merging:\ndst: ", dst);
}

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⭐ Possible output ⭐

Originally:
v1: 2 6 5 7 4 2 2 6 7 0
v2: 8 3 2 5 0 1 9 6 5 0
After sorting:
v1: 0 2 2 2 4 5 6 6 7 7
v2: 0 0 1 2 3 5 5 6 8 9
After merging:
dst: 0 0 0 1 2 2 2 2 3 4 5 5 5 6 6 6 7 7 8 9

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👑 Defect reports 👑

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

• LWG 780 — Applied to C++98 — Behavior as published: the merge operation was not defined — Correct behavior: defined.

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🫶 See also 🫶

• inplace_merge — merges two ordered ranges in-place (function template)
• is_sorted — checks whether a range is sorted into ascending order (C++11) (function template)
• set_union — computes the union of two sets (function template)
• sort — sorts a range into ascending order (function template)
• stable_sort — sorts a range of elements while preserving order between equal elements (function template)
• ranges::merge — merges two sorted ranges (C++20) (niebloid)

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okay that’s the whole page, bestie — source/version/date: http://cppreference.com, 2024.06.10, std::merge(3) 💖🎀✨