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std::ranges::set_union, std::ranges::set_union_result - cppreference.com

Call signature

Helper types

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 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.

1) Elements are compared using the given binary comparison function comp.

The function-like entities described on this page are algorithm function objects (informally known as niebloids), that is:

• Explicit template argument lists cannot be specified when calling any of them.
• None of them are visible to argument-dependent lookup.
• When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

{last1, last2, result_last}, where result_last is the end of the constructed range.

At most \(\scriptsize 2\cdot(N_1+N_2)-1\)2·(N1+N2)-1 comparisons and applications of each projection, where \(\scriptsize N_1\)N1 and \(\scriptsize N_2\)N2 are ranges::distance(first1, last1) and ranges::distance(first2, last2), respectively.

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 \(\scriptsize (N_1+N_2)\)(N1+N2) values and ranges::set_union may produce less.

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 {};

#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 }

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