Multiple iterators to a complex range

Question

I am trying to have multiple iterators to a bit more complex range (using range-v3 library) -- manually implementing a cartesian product, using filter, for_each and yield. However, when I tried to hold multiple iterators to such range, they share a common value. For example:

#include <vector>
#include <iostream>
#include <range/v3/view/for_each.hpp>
#include <range/v3/view/filter.hpp>

int main() {
    std::vector<int> data1{1,5,2,7,6};
    std::vector<int> data2{1,5,2,7,6};
    auto range =
            data1
            | ranges::v3::view::filter([](int v) { return v%2; })
            | ranges::v3::view::for_each([&data2](int v) {
                return data2 | ranges::v3::view::for_each([v](int v2) {
                    return ranges::v3::yield(std::make_pair(v,v2));
                });
            });
    auto it1 = range.begin();
    for (auto it2 = range.begin(); it2 != range.end(); ++it2) {
        std::cout << "[" << it1->first << "," << it1->second << "] [" << it2->first << "," << it2->second << "]\n";
    }
    return 0;
}

I expected the iterator it1 to keep pointing at the beginning of the range, while the iterator it2 goes through the whole sequence. To my surprise, it1 is incremented as well! I get the following output:

[1,1] [1,1]
[1,5] [1,5]
[1,2] [1,2]
[1,7] [1,7]
[1,6] [1,6]
[5,1] [5,1]
[5,5] [5,5]
[5,2] [5,2]
[5,7] [5,7]
[5,6] [5,6]
[7,1] [7,1]
[7,5] [7,5]
[7,2] [7,2]
[7,7] [7,7]
[7,6] [7,6]

• Why is that?
• How can I avoid this?
• How can I keep multiple, independent iterators pointing in various locations of the range?
• Should I implement a cartesian product in a different way? (that's my previous question)

---

While it is not reflected in the MCVE above, consider a use case where someone tries to implement something similar to std::max_element - trying to return an iterator to the highest-valued pair in the cross product. While looking for the highest value you need to store an iterator to the current best candidate. It cannot alter while you search, and it would be cumbersome to manage the iterators if you need a copy of the range (as suggested in one of the answers).

Materialising the whole cross product is not an option either, as it requires a lot of memory. After all, the whole point of using ranges with filters and other on-the-fly transformations is to avoid such materialisation.

Answer 1

It seems that the resulting view stores state such that it turns out to be single pass. You can work around that by simply making as many copies of the view as you need:

int main() {
    std::vector<int> data1{1,5,2,7,6};
    std::vector<int> data2{1,5,2,7,6};
    auto range =
            data1
            | ranges::v3::view::filter([](int v) { return v%2; })
            | ranges::v3::view::for_each([&data2](int v) {
                return data2 | ranges::v3::view::for_each([v](int v2) {
                    return ranges::v3::yield(std::make_pair(v,v2));
                });
            });

    auto range1= range;         // Copy the view adaptor
    auto it1 = range1.begin();

    for (auto it2 = range.begin(); it2 != range.end(); ++it2) {
        std::cout << "[" << it1->first << "," << it1->second << "] [" << it2->first << "," << it2->second << "]\n";
    }

    std::cout << '\n';
    for (; it1 != range1.end(); ++it1) { // Consume the copied view
        std::cout << "[" << it1->first << "," << it1->second << "]\n";
    }
    return 0;
}

Another option would be materializing the view into a container as mentioned in the comments.

---

Keeping in mind the aforementioned limitation of single-pass views, it is not really hard to implement a max_element function that returns an iterator, with the important drawback of having to compute the sequence one time and a half.

Here's a possible implementation:

template <typename InputRange,typename BinaryPred = std::greater<>>
auto my_max_element(InputRange &range1,BinaryPred &&pred = {}) -> decltype(range1.begin()) {
    auto range2 = range1;
    auto it1 = range1.begin();
    std::ptrdiff_t pos = 0L;

    for (auto it2 = range2.begin(); it2 != range2.end(); ++it2) {
        if (pred(*it2,*it1)) {
            ranges::advance(it1,pos);   // Computing again the sequence as the iterator advances!
            pos = 0L;
            }
        ++pos;
        }
    return it1; 
}