std::set_union

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

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

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

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

Constructs a hypothetical sorted union from two sorted source ranges [first1, last1) and [first2, last2), the union consists of the set of elements present at least one of the source ranges. Copies all elements of the union to the destination range beginning at d_first.

For each group of equivalent elements to be included in the union, let n1 and n2 be the numbers of elements from the two source ranges respectively:

• All n1 elements from the first source range will be included in the union in order.
• Then the first std::min(n1, n2) elements from the second source range will be skipped, and all remaining elments will be included in the union in order.

If the destination range overlaps with any of the two source ranges, the behavior is undefined.

Parameters

Return value

The past-the-end iterator of the destination range.

Complexity

Given

• \(\scriptsize N_1\)N1 as std::distance(first1, last1),
• \(\scriptsize N_2\)N2 as std::distance(first2, last2):

Exceptions

Notes

This algorithm performs a similar task as std::merge does. Both consume two sorted source ranges and produce a sorted output with elements from both inputs. The difference between these two algorithms is with handling values from both source ranges which compare equivalent (see notes on LessThanComparable).

If any equivalent values appeared n1 times in the first range and n2 times in the second, std::merge would output all n1 + n2 occurrences whereas std::set_union would output std::max(n1, n2) ones only. So std::merge outputs exactly std::distance(first1, last1) + std::distance(first2, last2) values and std::set_union may produce fewer.

Possible implementation

template<class InputIt1, class InputIt2, class OutputIt>
OutputIt set_union(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++;
        else
        {
            *d_first = *first1;
            if (!(*first1 < *first2))
                ++first2;
            ++first1;
        }
    }
    return std::copy(first2, last2, d_first);
}

template<class InputIt1, class InputIt2, class OutputIt, class Compare>
OutputIt set_union(InputIt1 first1, InputIt1 last1,
                   InputIt2 first2, InputIt2 last2, OutputIt d_first, Compare comp)
{
    for (; first1 != last1; ++d_first)
    {
        if (first2 == last2)
            // Finished range 2, include the rest of range 1:
            return std::copy(first1, last1, d_first);
        
        if (comp(*first2, *first1))
            *d_first = *first2++;
        else
        {
            *d_first = *first1;
            if (!comp(*first1, *first2)) // Equivalent => don't need to include *first2.
                ++first2;
            ++first1;
        }
    }
    // Finished range 1, include the rest of range 2:
    return std::copy(first2, last2, d_first);
}

Example

#include <algorithm>
#include <iterator>
#include <print>
#include <vector>

int main()
{
    std::vector<int> v1, v2, dest;
    
    v1 = {1, 2, 3, 4, 5};
    v2 = {3, 4, 5, 6, 7};
    
    std::set_union(v1.cbegin(), v1.cend(),
                   v2.cbegin(), v2.cend(),
                   std::back_inserter(dest));
    std::println("{}", dest);
    
    dest.clear();
    
    v1 = {1, 2, 3, 4, 5, 5, 5};
    v2 = {3, 4, 5, 6, 7};
    
    std::set_union(v1.cbegin(), v1.cend(),
                   v2.cbegin(), v2.cend(),
                   std::back_inserter(dest));
    std::println("{}", dest);
}

[1, 2, 3, 4, 5, 6, 7]
[1, 2, 3, 4, 5, 5, 5, 6, 7]

Defect reports

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

See also