std::ranges::set_union() algorithm

od C++20

Simplified

// (1)
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)
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 = {} );

The type of arguments are generic and have the following constraints:

• I1, I2 - std::input_iterator
• S1, S2 - std::sentinel_for<I1>, std::sentinel_for<I2>
• R1, R2 - std::ranges::input_range
• O - std::weakly_incrementable
• Comp - (none)
• Proj1, Proj2 - (none)

The Proj and Comp template arguments have the following default types: std::identity, ranges::less for all overloads.

Additionaly, each overload has the following constraints:

• (1) - mergeable<I1, I2, O, Comp, Proj1, Proj2>
• (2) - mergeable<ranges::iterator_t<R1>, ranges::iterator_t<R2>, O, Comp, Proj1, Proj2>

(The std:: namespace was ommited here for readability)

Detailed

// (1)
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)
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 = {} );

With the helper types defined as follows:

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

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).

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.

The order of equivalent elements is preserved.

• (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.

Undefined Behaviour

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.

The function-like entities described on this page are niebloids.

Parameters

first1 last1	The first sorted input range.
r1	The first sorted input range.
first2 last2	The second sorted input range.
r2	The second sorted input range.
result	The beginning of the destination range.
comp	Comparator 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.

Return value

A value of type ranges::set_union_result initialized as follows:

{
  last1,
  last2,
  result_last
}

Where result_last is the end of the constructed range.

Complexity

Given N1 as ranges::distance(first1, last1) and N2 as ranges::distance(first2, last12):

2 * (N1 + N2) − 1 comparisons and applications of each projection.

Exceptions

(none)

Possible implementation

set_union(1) and set_union(2)

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 {};

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.

Examples

Main.cpp

#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());
}

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 }