Iterate over equal ranges of range with C++20

Question

I am wondering if C++20 ranges have some nice way for me to iterate over the equal ranges of sorted container(or in general case any sorted range).

I have this "manual" solution that works, but it is not nice in a sense that it is not really composable (I do not get some view of equal range views back, I just provide function to be called).

#include <vector>
#include <ranges>
#include <iostream>
#include <fmt/ranges.h>


template<typename Cont, typename Fn>
void for_each_equal_range(const Cont& cont, Fn fn){
    auto current_begin = cont.begin();
    while(true){
        if (current_begin==cont.end()){
            return;
        }
        auto [eq_begin, eq_end] = std::equal_range(current_begin, cont.end(),   *current_begin);
        fn(eq_begin, eq_end);
        current_begin = eq_end;
    }
}

int main() {
    std::vector vals {1,2,2,3,3,3,47};
    for_each_equal_range(vals, [](const auto b, const auto e){
        std::cout << "size: " << std::distance(b,e) << std::endl;
        std::ranges::subrange elems(b, e);
        std::cout << fmt::format("{}",elems) << std::endl;
    });
}

I wish I had something like:

vals | equal_range_split | std::ranges::for_each(...);

In case there is some confusion wrt what range means here:

equal range is good old STL equal_range meaning
ranges is C++20 ranges library.

Also I know C++20 has std::ranges::equal_range algorithm, but it seems to be not that helpful for my use case.

You're looking for `group_by`, but that's not in C++20. I take it you don't want a range-v3 solution? — cigien, Jul 20 '21 at 01:56
@cigien no, not really, kind of hard to justify use of range-v3 in real code just for this little functionality... and few others like cartesian product, ... — NoSenseEtAl, Jul 20 '21 at 02:09
Yeah, then you'll just have to write it yourself for now. It may be in the pipeline for C++23, but I'm not sure. — cigien, Jul 20 '21 at 02:11
@cigen if you make that an answer I can accept it. It is not great news, but it answers the question. — NoSenseEtAl, Jul 20 '21 at 02:42

Caleth · Answer 1 · 2021-07-21T08:10:49.660

Not quite what you are looking for, but if you have a coroutine generator template, something like

using namespace std::ranges;

template<typename Range, typename Compare = std::less>
generator<subrange<iterator_t<Range>>> equal_ranges(Range&& range, Compare compare = {}) {
    for (auto current_begin = cont.begin(); current_begin != cont.end();) {
        auto [eq_begin, eq_end] = std::equal_range(current_begin, cont.end(), *current_begin, compare);
        co_yield { eq_begin, eq_end };
        current_begin = eq_end;
    }
}

template<typename Compare>
struct equal_ranges_holder {
    Compare compare;
};

template<typename Compare = std::less>
equal_ranges_holder<Compare> equal_ranges(Compare compare = {}) { return { compare }; } 

template<typename Range, typename Compare>
auto operator|(Range&& range, equal_ranges_holder<Compare> holder) {
    return equal_ranges(range, holder.compare);
}

Alternatively, it's fairly simple to make a single pass equal_range_view.

using namespace std::ranges;

template<borrowed_range Range, std::strict_weak_order Compare = std::less>
class equal_range_view : public view_interface<equal_range_view> {
    using base_iterator = iterator_t<Range>;
    using base_sentinel = sentinel_t<Range>;

    Range range;
    Compare compare;
public:
    class sentinel {};

    class iterator {
        base_iterator base;
        base_sentinel end;
        Compare compare;
    public:
        using value_type = subrange<base_iterator>;
        using reference = value_type;
        using pointer = value_type *;
        using difference_type = std::ptrdiff_t;
        using iterator_category = std::input_iterator_tag;
    
        equal_range_iterator(base_iterator base, base_sentinel end, Compare compare) : base(base), end(end), compare(compare) {}

        reference operator*() {
            auto [first, last] = std::equal_range(base, end, compare);
            return { first, last };
        }
    
        iterator& operator++() {
            auto [_, last] = std::equal_range(base, end, compare);
            base = last;
            return *this;
        }

        bool operator==(sentinel) const {
            return base == end;
        }
        bool operator!=(sentinel) const {
            return base != end;
        }
    }; 

    explicit equal_range_view(Range range, Compare compare = {}) : range(range), compare(compare) {}

    iterator begin() { return { range.begin(), range.end(), compare }; }
    sentinel end() { return {}; }
};

Hi, is this generator something you wrote, or you are using some open source library(till C++ gets generators in std::) ? — NoSenseEtAl, Jul 20 '21 at 14:14

score 1 · Answer 2 · answered Jul 20 '21 at 07:08

1

In C++20, pipe operator has not defined for std::ranges::equal_range and std::ranges::for_each yet. So simple technique to implement above code is like:

  for (int global = *vals.begin() - 1 ; 
           auto val : vals | std::ranges::views::filter([&](auto value){return value != global ;}))
    {
        std::ranges::for_each(std::ranges::equal_range(vals, val), [](auto& x){ std::cout << x << std::endl; });
        global = val;
    }

answered Jul 20 '21 at 07:08

AmirSalar

325
2
14

underflowing an int is undefined behavior(assume vals[0] is INT_MIN) – NoSenseEtAl Jul 20 '21 at 07:12
1

Your code will exhibit UB, because your predicate is **not** a [`regular_invocable`](http://eel.is/c++draft/concepts#concept:regular_invocable). – 康桓瑋 Jul 20 '21 at 07:12
@NoSenseEtAl the `global` variable is set to anything except `*vals.begin()`, so it could be set `*vals.begin() + 1`. but I agree that this variable has the potential to sabotage. – AmirSalar Jul 20 '21 at 07:39

Iterate over equal ranges of range with C++20

2 Answers2