std::set_union (3) - Linux Manuals
std::set_union: std::set_union
NAME
std::set_union - std::set_union
Synopsis
Defined in header <algorithm>
template< class InputIt1, class InputIt2, class OutputIt >
OutputIt set_union( InputIt1 first1, InputIt1 last1, (until C++20)
InputIt2 first2, InputIt2 last2,
OutputIt d_first );
template< class InputIt1, class InputIt2, class OutputIt >
constexpr OutputIt set_union( InputIt1 first1, InputIt1 last1, (since C++20)
InputIt2 first2, InputIt2 last2,
OutputIt d_first );
template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2, class ForwardIt3 >
ForwardIt3 set_union( ExecutionPolicy&& policy, ForwardIt1 first1, ForwardIt1 last1, (2) (since C++17)
ForwardIt2 first2, ForwardIt2 last2,
ForwardIt3 d_first );
template< class InputIt1, class InputIt2,
class OutputIt, class Compare > (1)
OutputIt set_union( InputIt1 first1, InputIt1 last1, (until C++20)
InputIt2 first2, InputIt2 last2,
OutputIt d_first, Compare comp );
template< class InputIt1, class InputIt2,
class OutputIt, class Compare >
constexpr OutputIt set_union( InputIt1 first1, InputIt1 last1, (3) (since C++20)
InputIt2 first2, InputIt2 last2,
OutputIt d_first, Compare comp );
template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2,
class ForwardIt3, class Compare >
ForwardIt3 set_union( ExecutionPolicy&& policy, ForwardIt1 first1, ForwardIt1 last1, (4) (since C++17)
ForwardIt2 first2, ForwardIt2 last2,
ForwardIt3 d_first, Compare comp );
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 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 d_first, preserving order, and then exactly std::max(n-m, 0) elements will be copied from [first2, last2) to d_first, also preserving order.
The resulting range cannot overlap with either of the input ranges.
1) Elements are compared using operator< and the ranges must be sorted with respect to the same.
3) Elements are compared using the given binary comparison function comp and the ranges must be sorted with respect to the same.
2,4) Same as (1,3), but executed according to policy. This overload only participates in overload resolution if std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> is true
Parameters
first1, last1 - the first input sorted range
first2, last2 - the second input sorted range
policy - the execution policy to use. See execution_policy for details.
comp - 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
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.
Return value
Iterator past the end of the constructed range.
Complexity
At most 2ยท(N1+N2-1) comparisons, where N1 = std::distance(first1, last1) and N2 = std::distance(first2, last2).
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.
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 less.
Possible implementation
First version