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I have seen some similar questions, like this, this and this, however they all are quite old and might be obsolete.

It is 2023 and the latest C++ standard is C++20 which was published in 2020, and different standards of C++ are very different.

So in C++20 are there library methods for set difference, intersection and union of unordered_set that was included using #include <unordered_set>?

I know in Python, if a and b are sets, a & b is the intersection, a - b is one way difference (elements in a that are not in b), a ^ b is the symmetric difference, a | b is the union. Are there similar methods in C++ that are more efficient than for loop based solutions?

I am reading this and there seems to be set_union, set_difference and set_intersection methods, but I don't know what those are.

Of course I know how to brute force it:

#include <iostream> #include <unordered_set> using intSet = std::unordered_set<int>; void print_set(const intSet& set) { std::cout << "{"; for (int i : set) { std::cout << i << ", "; } std::cout << "}\n"; } int main() { intSet a = { 16, 1, 14, 7, 18, 5, 12, 19 }; intSet b = { 0, 9, 1, 8, 19, 2, 18, 13 }; intSet intersection_set; intSet diff_set; intSet union_set; for (int i : a) { if (b.count(i)) { intersection_set.insert(i); } else { diff_set.insert(i); } union_set.insert(i); } for (int i : b) { union_set.insert(i); } std::cout << "a = "; print_set(a); std::cout << "b = "; print_set(b); std::cout << "a & b = "; print_set(intersection_set); std::cout << "a - b = "; print_set(diff_set); std::cout << "a | b = "; print_set(union_set); } 
a = {16, 1, 14, 7, 18, 5, 12, 19, } b = {8, 0, 1, 9, 19, 18, 2, 13, } a & b = {1, 18, 19, } a - b = {16, 14, 7, 5, 12, } a | b = {16, 1, 14, 7, 18, 5, 12, 19, 8, 0, 9, 2, 13, }
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    The Python article just seems wrong, as in C++ an unordered_set is very different from a set. The set algorithms have a precondition that the sets are ordered, which the unordered_set obviously is not. Commented Aug 25, 2023 at 9:52
  • You say, more efficient than for loop based solutions ... but what makes you think, if such methods were implemented for unordered sets, that they wouldn't be (of necessity) loop-based? Commented Aug 25, 2023 at 11:08
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    You won't get much better than your "brute force" method. For the intersection, you can iterate over the smaller set, but that is all you can do when dealing with unordered sets. Commented Aug 25, 2023 at 11:14
  • Maybe you could implement &, | and others for unordered (and even ordered) sets and offer them to the Standards Committee for release in C++26? Commented Aug 25, 2023 at 11:22

1 Answer 1

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The algorithms from <algorithm> aren't designed to work with std::unordered_set but with ordered ranges. For std::unordered_set, you can use its member functions in order to compute unions and differences, and std::erase_if for intersections.

In-place

#include <unordered_set> std::unordered_set<int> a, b; // TODO: fill with data // a | b (set union) a.insert(b.begin(), b.end()); // better alternative if you are allowed to modify a AND b a.merge(b); // a - b (set difference) a.erase(b.begin(), b.end()); // a & b (set intersection) std::erase_if(a, [&b](int e) { return !b.contains(e); });

Non-modifying

#include <unordered_set> #include <ranges> const std::unordered_set<int> a, b; // TODO: fill with data // a | b (set union) auto ab = {a, b}; auto joined = ab | std::views::join; std::unordered_set<int> u(joined.begin(), joined.end()); // a - b (set difference) auto diff = a | std::views::filter([&b](int e) { return !b.contains(e); }); std::unordered_set<int> d(diff.begin(), diff.end()); // a & b (set intersection) auto isect = a | std::views::filter([&b](int e) { return b.contains(e); }); std::unordered_set<int> i(isect.begin(), isect.end());

Note: these declarations can be simplified in C++23 with std::from_range.

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