std::set_intersection(3)

💖 NAME 💖

Okay so bestie, this is <pink>std::set_intersection(3)</pink> from the <purple>C++ Standard Libary</purple> — yes, the manual name is exactly std::set_intersection(3), and the thing itself is std::set_intersection. It's giving intersection-algorithm main character energy. 💅

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✨ Synopsis ✨

Defined in header <algorithm>:

template< class InputIt1, class InputIt2, class OutputIt >
OutputIt set_intersection( 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_intersection( 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_intersection( 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_intersection( ExecutionPolicy&& policy,
                             ForwardIt1 first1, ForwardIt1 last1,
                             ForwardIt2 first2, ForwardIt2 last2,
                             ForwardIt3 d_first, Compare comp ); (4) (since C++17)

Constructs a sorted range beginning at d_first consisting of elements that are found in 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, the first std::min(m, n) elements will be copied from [first1, last1) to the output range, preserving order.

• 1) If [first1, last1) or [first2, last2) is not sorted with respect to operator< (until C++20) or 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. 💀

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🌸 Parameters 🌸

• first1, last1 — the first range of elements to examine.
• first2, last2 — the second range of elements to examine.
• 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.

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🦋 Type requirements 🦋

• InputIt1, InputIt2 must meet the requirements of LegacyInputIterator.
• OutputIt must meet the requirements of LegacyOutputIterator.
• ForwardIt1, ForwardIt2, ForwardIt3 must meet the requirements of LegacyForwardIterator.
• Compare must meet the requirements of Compare.

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👑 Return value 👑

Iterator past the end of the constructed range.

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🔥 Complexity 🔥

Given N_1 as std::distance(first1, last1) and N_2 as std::distance(first2, last2):

• 1,2) At most 2 * (N_1 + N_2) - 1 comparisons using operator< (until C++20) or std::less{} (since C++20).
• 3,4) At most 2 * (N_1 + N_2) - 1 applications of the comparison function comp.

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💀 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.

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🧚 Possible implementation 🧚

set_intersection (1)

template<class InputIt1, class InputIt2, class OutputIt>
OutputIt set_intersection(InputIt1 first1, InputIt1 last1,
                          InputIt2 first2, InputIt2 last2,
                          OutputIt d_first)
{
    while (first1 != last1 && first2 != last2)
    {
        if (*first1 < *first2)
            ++first1;
        else
        {
            if (!(*first2 < *first1))
                *d_first++ = *first1++; // *first1 and *first2 are equivalent.
            ++first2;
        }
    }
    return d_first;
}

set_intersection (3)

template<class InputIt1, class InputIt2, class OutputIt, class Compare>
OutputIt set_intersection(InputIt1 first1, InputIt1 last1,
                          InputIt2 first2, InputIt2 last2,
                          OutputIt d_first, Compare comp)
{
    while (first1 != last1 && first2 != last2)
    {
        if (comp(*first1, *first2))
            ++first1;
        else
        {
            if (!comp(*first2, *first1))
                *d_first++ = *first1++; // *first1 and *first2 are equivalent.
            ++first2;
        }
    }
    return d_first;
}

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🌈 Example 🌈

Run this code:

// Run this code

#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>

int main()
{
    std::vector<int> v1{7, 2, 3, 4, 5, 6, 7, 8};
    std::vector<int> v2{5, 7, 9, 7};
    std::sort(v1.begin(), v1.end());
    std::sort(v2.begin(), v2.end());

    std::vector<int> v_intersection;
    std::set_intersection(v1.begin(), v1.end(), v2.begin(), v2.end(),
                          std::back_inserter(v_intersection));

    for (int n : v_intersection)
        std::cout << n << ' ';
    std::cout << '\n';
}

🖥️ Output 🖥️

5 7 7

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🎀 Defect reports 🎀

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

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

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⭐ See also ⭐

• set_union — computes the union of two sets (function template)
• ranges::set_intersection — computes the intersection of two sets (C++20) (niebloid)

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Okay bestie, that's the whole page! 💖 Cute little sign-off with the source/version/date: http://cppreference.com, 2024.06.10, page <pink>std::set_intersection(3)</pink>. Slay. 🫶