How to use output iterators without a container: Set intersection without storage

Question

The C++ reference has this example for printing the intersection of two sets (or sorted containers):

#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';
}

In my application, I simply want to iterate through the intersection of two sets; I do not need to store that intersection.

Question: What is the cleanest way to iterate through the intersection of two sets without storing the intersection?

By "cleanest" I mean without re-implementing set_intersection. Here is what I use; this "hacks" the inserter to call a function of my choosing:

#include <stack>
#include <set>
#include <vector>
#include <string>
#include <iostream>

template <typename F>
struct output_function {
    using value_type = int;
    output_function (F f) : f (f) {}
    F f;
    void push_back (int x) { f (x); }
};

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());

  int z = 0;
  auto f = [&z] (int t) { std::cout << z++ << ": " << t << "\n"; };
  output_function p (f);

  std::set_intersection(v1.begin(), v1.end(), v2.begin(), v2.end(),
                        std::back_inserter (p));
}

I would make something like an executing_iterator which would be a class type that mimics being an iterator but when you write to it, it passes the values given into a function that you pass to the iterator when you create it. — NathanOliver
– NathanOliver, Commented Apr 20 at 16:34
What you've got looks pretty clean to me but I'd probably make a separate function for it anyway. It's not much more code than the class you've added and it's probably clearer for a reader. — Ted Lyngmo
– Ted Lyngmo, Commented Apr 20 at 16:51

Sebastian Redl · Accepted Answer · 2025-04-20 16:46:19Z

5

If you are willing to use Boost, you can use function_output_iterator.

It's mostly straight-forward to implement yourself, but the logic of having an assignment operator that takes an arbitrary type without hiding the copy assignment operator is a bit annoying.

answered Apr 20 at 16:46

Sebastian Redl

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2 Comments

Michaël Apr 20 at 22:21

Aah, very nice! Any clue why it works well with std::set_intersection but not with std::ranges::set_intersection? godbolt.org/z/KGv9v718K

Sebastian Redl Apr 21 at 14:15

The ranges version validates the concepts, whether the set_intersection implementation uses them or not. And it requires that the output iterator be weakly_incrementable, which in turns requires movable, which requires assignable_from<T&, T>, i.e. the output iterator must be assignable. But the iterator contains a lambda which has a by-reference capture, and that's not assignable.

康桓瑋 · Accepted Answer · 2025-04-21 02:14:57Z

5

You can use range-v3's views::set_intersection:

#include <algorithm>
#include <vector>
#include <range/v3/view/set_algorithm.hpp>
 
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());
 
    auto intersection = ranges::views::set_intersection(v1, v2);
    for (int n : intersection)
      // ...
}

Demo

answered Apr 21 at 2:14

康桓瑋

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Comments

Dominik Kaszewski · Accepted Answer · 2025-04-20 20:50:09Z

0

This should get you most of the way there by mimicking implementation of std::insert_iterator

// TODO: Convert to class, add CTAD
template <typename T>
struct function_iterator {
    // TODO: iterator_traits stuff required by algorithms
    // TODO: Store a functor instead
    std::function<void(const T&)> fun{};

    void operator++() {}
    void operator++(int) {}
    auto& operator*() { return *this; }
    // TODO: Move overload
    auto& operator=(const T& val) { fun(val); return *this; }
};

edited Apr 20 at 20:50

answered Apr 20 at 19:59

Dominik Kaszewski

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