tmp/tmpkcu9v1j1/{from.md → to.md}
RENAMED
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#### `set_union` <a id="set.union">[[set.union]]</a>
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``` cpp
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template<class InputIterator1, class InputIterator2,
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class OutputIterator>
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OutputIterator
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set_union(InputIterator1 first1, InputIterator1 last1,
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InputIterator2 first2, InputIterator2 last2,
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OutputIterator result);
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template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2,
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class ForwardIterator>
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@@ -15,37 +15,60 @@ template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2,
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ForwardIterator2 first2, ForwardIterator2 last2,
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ForwardIterator result);
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template<class InputIterator1, class InputIterator2,
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class OutputIterator, class Compare>
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OutputIterator
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set_union(InputIterator1 first1, InputIterator1 last1,
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InputIterator2 first2, InputIterator2 last2,
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OutputIterator result, Compare comp);
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template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2,
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class ForwardIterator, class Compare>
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ForwardIterator
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set_union(ExecutionPolicy&& exec,
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ForwardIterator1 first1, ForwardIterator1 last1,
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ForwardIterator2 first2, ForwardIterator2 last2,
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ForwardIterator result, Compare comp);
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```
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-
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original ranges.
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*Effects:* Constructs a sorted union of the elements from the two
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ranges; that is, the set of elements that are present in one or both of
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the ranges.
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*Returns:*
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*Complexity:* At most `2 * ((last1 - first1) + (last2 - first2)) - 1`
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comparisons.
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*Remarks:* If \[`first1`, `last1`) contains
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equivalent to each other and \[`first2`, `last2`)
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that are equivalent to them, then all m elements
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-
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elements from the second range
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order.
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#### `set_union` <a id="set.union">[[set.union]]</a>
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``` cpp
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template<class InputIterator1, class InputIterator2,
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class OutputIterator>
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+
constexpr OutputIterator
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set_union(InputIterator1 first1, InputIterator1 last1,
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InputIterator2 first2, InputIterator2 last2,
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OutputIterator result);
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template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2,
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class ForwardIterator>
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ForwardIterator2 first2, ForwardIterator2 last2,
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ForwardIterator result);
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template<class InputIterator1, class InputIterator2,
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class OutputIterator, class Compare>
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constexpr OutputIterator
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set_union(InputIterator1 first1, InputIterator1 last1,
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InputIterator2 first2, InputIterator2 last2,
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OutputIterator result, Compare comp);
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template<class ExecutionPolicy, class ForwardIterator1, class ForwardIterator2,
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class ForwardIterator, class Compare>
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ForwardIterator
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set_union(ExecutionPolicy&& exec,
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ForwardIterator1 first1, ForwardIterator1 last1,
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ForwardIterator2 first2, ForwardIterator2 last2,
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ForwardIterator result, Compare comp);
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+
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template<input_iterator I1, sentinel_for<I1> S1, input_iterator I2, sentinel_for<I2> S2,
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weakly_incrementable O, class Comp = ranges::less,
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class Proj1 = identity, class Proj2 = identity>
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requires mergeable<I1, I2, O, Comp, Proj1, Proj2>
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constexpr ranges::set_union_result<I1, I2, O>
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ranges::set_union(I1 first1, S1 last1, I2 first2, S2 last2, O result, Comp comp = {},
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Proj1 proj1 = {}, Proj2 proj2 = {});
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template<input_range R1, input_range R2, weakly_incrementable O,
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class Comp = ranges::less, class Proj1 = identity, class Proj2 = identity>
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requires mergeable<iterator_t<R1>, iterator_t<R2>, O, Comp, Proj1, Proj2>
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constexpr ranges::set_union_result<borrowed_iterator_t<R1>, borrowed_iterator_t<R2>, O>
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ranges::set_union(R1&& r1, R2&& r2, O result, Comp comp = {},
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Proj1 proj1 = {}, Proj2 proj2 = {});
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```
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Let `comp` be `less{}`, and `proj1` and `proj2` be `identity{}` for the
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overloads with no parameters by those names.
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*Preconditions:* The ranges \[`first1`, `last1`) and \[`first2`,
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`last2`) are sorted with respect to `comp` and `proj1` or `proj2`,
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respectively. The resulting range does not overlap with either of the
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original ranges.
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*Effects:* Constructs a sorted union of the elements from the two
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ranges; that is, the set of elements that are present in one or both of
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the ranges.
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*Returns:* Let `result_last` be the end of the constructed range.
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Returns
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- `result_last` for the overloads in namespace `std`.
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- `{last1, last2, result_last}` for the overloads in namespace `ranges`.
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*Complexity:* At most `2 * ((last1 - first1) + (last2 - first2)) - 1`
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comparisons and applications of each projection.
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*Remarks:* Stable [[algorithm.stable]]. If \[`first1`, `last1`) contains
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m elements that are equivalent to each other and \[`first2`, `last2`)
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contains n elements that are equivalent to them, then all m elements
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from the first range are copied to the output range, in order, and then
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the final max(n - m, 0) elements from the second range are copied to the
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output range, in order.
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