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std::set_union - cppreference.com

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.

If

or

is not

with respect to

, 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 all following conditions are satisfied:

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

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. ​

Iterator past the end of the constructed range.

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

At most

comparisons using

.

3,4) At most \(\scriptsize 2 \cdot (N_1+N_2)-1\)2⋅(N1+N2)-1 applications of the comparison function comp.

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.

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);
}

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.

#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

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

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