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

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.

The behavior is undefined if the destination range overlaps either of the input ranges (the input ranges may overlap each other).

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 algorithm function objects (informally known as 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.

[edit] Parameters

[edit] Return value

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

[edit] Complexity

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

[edit] 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.

[edit] 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 {};

[edit] Example

#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());
}

{ 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 }

[edit] See also

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