std::set_union

Defined in header `<algorithm>`
template< class InputIt1, class InputIt2, class OutputIt > OutputIt set_union( InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first );	(1)	(constexpr since C++20)
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 );	(2)	(since C++17)
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 );	(3)	(constexpr since C++20)
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 );	(4)	(since C++17)

Constructs a sorted union beginning at d_first consisting of the set of elements present in one or both sorted ranges [first1, last1) and [first2, last2).

If [first1, last1) contains m elements that are equivalent to each other and [first2, last2) contains n elements that are equivalent to them, then all m elements will be copied from [first1, last1) to the output range, preserving order, and then the final std::max(n - m, 0) elements will be copied from [first2, last2) to the output range, also preserving order.

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)

If the output range overlaps with [first1, last1) or [first2, last2), the behavior is undefined.

Parameters

first1, last1	-	the first input sorted range
first2, last2	-	the second input sorted range
d_first	-	the beginning of the output 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.
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.

Return value

Iterator past the end of the constructed range.

Complexity

Given $\scriptsize N_1$ $N 1$ as std::distance(first1, last1) and $\scriptsize N_2$ $N 2$ as std::distance(first2, last2):

1,2) At most

2\cdot(N 1 +N 2)-1

comparisons using operator<(until C++20)std::less{}(since C++20).

3,4) At most

2\cdot(N 1 +N 2)-1

applications of the comparison function comp.

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.

Possible implementation

set_union (1)

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);
}

set_union (3)

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);
}

Notes

This algorithm performs a similar task as std::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, 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.

Example

Run this code

#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
 
void println(const std::vector<int>& v)
{
    for (int i : v)
        std::cout << i << ' ';
    std::cout << '\n';
}
 
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));
    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));
    println(dest);
}

Output:

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.

DR	Applied to	Behavior as published	Correct behavior
LWG 291	C++98	it was unspecified how to handle equivalent elements in the input ranges	specified

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External libraries

includes	returns true if one sequence is a subsequence of another (function template)
merge	merges two sorted ranges (function template)
set_difference	computes the difference between two sets (function template)
set_intersection	computes the intersection of two sets (function template)
set_symmetric_difference	computes the symmetric difference between two sets (function template)
ranges::set_union (C++20)	computes the union of two sets (niebloid)