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# std::set_intersection

From cppreference.com
< cpp‎ | algorithm

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) |
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 ); | (2) |

Constructs a sorted range beginning at `d_first` consisting of elements that are found in both sorted ranges `[first1, last1)` and `[first2, last2)`. The first version expects both input ranges to be sorted with operator<, the second version expects them to be sorted with the given comparison function `comp`. 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.

### [edit] Parameters

first1, last1 | - | the first range of elements to examine
first2, last2 | - | the second range of elements to examine
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); The signature does not need to have const &, but the function object must not modify the objects passed to it. The types Type1 and Type2 must be such that objects of types InputIt1 and InputIt2 can be dereferenced and then implicitly converted to Type1 and Type2 respectively. ​
Type requirements
- `InputIt1` must meet the requirements of `InputIterator`.
- `InputIt2` must meet the requirements of `InputIterator`.
- `OutputIt` must meet the requirements of `OutputIterator`.

### [edit] Return value

Iterator past the end of the constructed range.

### [edit] Complexity

At most 2·(N1+N2-1) comparisons, where N1 = std::distance(first1, last1) and N2 = std::distance(first2, last2).

### [edit] 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++;
                }
                ++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++;
                }
                ++first2;
            }
        }
        return d_first;
    }

### [edit] Example

Run this code

    #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

### [edit] See also

set_union | computes the union of two sets (function template) [edit]

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