size() causes error when switching from transform to filter

When

• views::filter([](auto n) { return n % 2 == 0; }); is activated, I fail to get [2,4].
• views::transform([](auto n) { return 2 * n; }); is activated, I succeed to get [2,4,6,8,10].

<!---->

#include <iostream>
#include <ranges>
#include <sstream>

int main()
{
    using namespace std;
    auto input = views::iota(1, 5 + 1);

    auto output = input |
                  views::filter([](auto n) { return n % 2 == 0; });
                  //views::transform([](auto n) { return 2 * n; });


    stringstream ss;
    ss << "[";
    for (auto i = 0; i < output.size() - 1; ++i)
    {
        ss << output[i] << ',';
    }
    ss << output[output.size() - 1];
    ss << "]";
    cout << ss.str();
    return 0;
}

I don't understand the following error.

> Filtering.cpp:17:37: error: no matching function for call to 'std::ranges::filter_view<std::ranges::iota_view<int, int>, main()::<lambda(auto:17)> >::size()'
17 | for (auto i = 0; i < output.size() - 1; ++i)

std::ranges::filter_view is not a "sized range", which means you can't call size on it. This is because the filtering is done on-demand, so asking for the size of the view would require consuming every element.

One option is to convert the view into a std::vector and use that. In C++20:

auto output_view = input
            | views::filter([](auto n) { return n % 2 == 0; }) 
            | views::transform([](auto n) { return 2 * n; });

std::vector<int> output(ranges::begin(output_view), ranges::end(output_view));

In C++23:

auto output = input
            | views::filter([](auto n) { return n % 2 == 0; }) 
            | views::transform([](auto n) { return 2 * n; })
            | ranges::to<std::vector>();

Another option is to make use of the fact that filter_view is a forward range so long as the underlying range is, so you can check if advancing the iterator hits the end:

stringstream ss;
ss << "[";
auto it = ranges::begin(output);
for (; ranges::next(it) != ranges::end(output); ++it) {
    ss << *it << ',';
}
ss << *it;
ss << "]";
cout << ss.str();

A filter_view is not a random_access_range and it does not have a size() method. If you focus your attention on the exact error message that you showed that's what it says: there is no size(), anywhere in the vicinity.

You can think of it as analogous to a std::forward_list, instead of std::vector. You can use range iteration, instead:

    const char *sep="";
    for (const auto &n:output)
    {
	    ss << sep << n;
	    sep=",";
    }

#include <iostream>
#include <ranges>
#include <sstream>


int main()
{
    using namespace std;
    auto input = views::iota(1, 5 + 1);

    auto output = input | views::filter([](const auto& n) {return n % 2 == 0; });

    stringstream ss;
    ss << "[";
    if (not ranges::empty(output))
    {
        ss << *ranges::begin(output);
        for (const auto &x : output | views::drop(1)) 
            ss << ',' << x;
    }
    ss << "]";
    cout << ss.str();
    return 0;
}