c++20 ranges view to vector

Question

Now that the C++20 ranges implementation is actually here and released under GCC 10.2, I would like to know how to convert a ranges view back to an actual container, like a vector. I've found this question ( Range view to std::vector ) that asked the same thing for the pre-release version but I would like to know if since its been released, have there been any new methods made to convert from a view to a container? Or is that single answer on that question still the best solution?

Answer 1

Easiest thing to do would be to use range-v3, which has a conversion operator exactly for this. From the examples:

using namespace ranges;
auto vi =
    views::for_each(views::ints(1, 10), [](int i) {
        return yield_from(views::repeat_n(i, i));
    })
  | to<std::vector>();
// vi == {1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...}

---

Otherwise, the answer in the linked question isn't entirely accurate since a range may not have the same iterator and sentinel types, and the answer requires it. So we can do a little bit better:

template <std::ranges::range R>
auto to_vector(R&& r) {
    std::vector<std::ranges::range_value_t<R>> v;

    // if we can get a size, reserve that much
    if constexpr (requires { std::ranges::size(r); }) {
        v.reserve(std::ranges::size(r));
    }

    // push all the elements
    for (auto&& e : r) {
        v.push_back(static_cast<decltype(e)&&>(e));
    }

    return v;
}

A shorter version of the above, which doesn't necessarily reserve up front in all the same places, addresses the mixed-sentinel-type issue by using views::common:

template <std::ranges::range R>
auto to_vector(R&& r) {
    auto r_common = r | std::views::common;
    return std::vector(r_common.begin(), r_common.end());
}

The canonical example of missed reserve here would be invoking to_vector() with a std::list<T> - which has an O(1) size() available, which could be used to reserve, but we lose that once we go into the iterators.