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std::set_intersection

Defined in header <algorithm>
(1)
template< class InputIt1, class InputIt2, class OutputIt >
OutputIt set_intersection( InputIt1 first1, InputIt1 last1,
                           InputIt2 first2, InputIt2 last2,
                           OutputIt d_first );
(until C++20)
template< class InputIt1, class InputIt2, class OutputIt >
constexpr OutputIt set_intersection( InputIt1 first1, InputIt1 last1,
                                     InputIt2 first2, InputIt2 last2,
                                     OutputIt d_first );
(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)
(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 );
(until C++20)
template< class InputIt1, class InputIt2, class OutputIt, class Compare >
constexpr OutputIt set_intersection( InputIt1 first1, InputIt1 last1,
                                     InputIt2 first2, InputIt2 last2,
                                     OutputIt d_first, Compare comp );
(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 some element is found m times in [first1, last1) and n times in [first2, last2), the first std::min(m, n) elements will be copied from the first range to the destination range. The order of equivalent elements is preserved. 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. These overloads do not participate in overload resolution unless std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> (until C++20) std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>> (since C++20) is true.

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 destination 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. ​

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.

Return value

Iterator past the end of the constructed range.

Complexity

At most \(\scriptsize 2\cdot(N_1+N_2)-1\)2·(N
1
+N
2
)-1 comparisons, where \(\scriptsize N_1\)N
1
and \(\scriptsize N_2\)N
2
are std::distance(first1, last1) and std::distance(first2, last2), respectively.

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.

Possible implementation

First version
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;
}
Second version
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;
}

Example

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
int main()
{
    std::vector<int> v1{1,2,3,4,5,6,7,8};
    std::vector<int> v2{        5,  7,  9,10};
    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 << ' ';
}

Output:

5 7

See also

computes the union of two sets
(function template)
(C++20)
computes the intersection of two sets
(niebloid)

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