Range view to std::vector

In the proposed C++20 (The One) Ranges TS, what is the proposed method for converting the view into a std::vector?

The following code does not compile:

int                                           
main() {                                                        
    std::vector<float> values = {1.0, 2.0, 3.0, 4.0, 5.2, 6.0, 7.0, 8.0, 9.0}; 
    //fmt::print("{}\n", std::experimental::ranges::views::filter(values, [] (float v) { return v < 5.f; })); 
    std::vector<float> foo = vw::filter(values, [] (float v) { return v < 5.f; }); 
    fmt::print("{}\n", foo);                
}

with the error

../src/view.cpp:19:40: error: conversion from     ‘std::experimental::ranges::v1::filter_view<std::experimental::ranges::v1::ref_view<std::vector<float> >, main()::<lambda(float)> >’ to non-scalar type ‘std::vector<float>’ requested
     std::vector<float> foo = vw::filter(values, [] (float v) { return v < 5.f; });

(the commented line will also not compile due to some CV constraints).

So how do I do anything with a view except for using a range-based for loop?

Also some bonus questions:

Is the cmcstl2 implementation I used even following the proposal? The ranges-v3 seems not to be.
Is there any documentation on the Ranges TS? The proposal PDF I found is pretty much an awfully formatted code dump in diff style. In fact directly reading the cmcstl2 sources was way easier to read for me. The cppreference seems to be lacking as well...

Solution

The C++20 method to convert a view to a std::vector (or indeed any other container) is to pass the range's begin and end members to the vector constructor that accepts 2 iterators (and an optional allocator).

I was also looking for an answer to this question. What I really wanted is an overload of the constructor of std::vector accepting a range. Approximately:

template <std::ranges::input_range R>
vector(R&& r) : vector(r.begin(), r.end()) {
}

but that isn't in C++20.

First, I implemented this:

namespace rng = std::ranges;

template <rng::range R>
constexpr auto to_vector(R&& r) {
    using elem_t = std::decay_t<rng::range_value_t<R>>;
    return std::vector<elem_t>{r.begin(), r.end()};
}

which works, but isn't very "rangy": https://godbolt.org/z/f2xAcd

I then did it a bit better:

namespace detail {
    // Type acts as a tag to find the correct operator| overload
    template <typename C>
    struct to_helper {
    };
    
    // This actually does the work
    template <typename Container, rng::range R>
    requires std::convertible_to<rng::range_value_t<R>, typename Container::value_type>
    Container operator|(R&& r, to_helper<Container>) {
        return Container{r.begin(), r.end()};
    }
}

// Couldn't find an concept for container, however a
// container is a range, but not a view.
template <rng::range Container>
requires (!rng::view<Container>)
auto to() {
    return detail::to_helper<Container>{};
}

https://godbolt.org/z/G8cEGqeq6

No doubt one can do better for sized_ranges and containers like std::vector that have a reserve member function.

C++23 has added a to<Container> function - https://en.cppreference.com/w/cpp/ranges/to. You can use a template (of one type parameter) or a concrete type as the parameter. E.g.: to::<std::vector<int>> vs to::<std::vector>. In the latter case the type used to instantiate the container is deduced so attention will have to be paid that it turns out as desired.

Note that the standard requires that for containers that have reserve and append (i.e. push_back/emplace_back/etc) functionality the implementation must do the efficient thing of calling reserve first.