[SOLVED] How to split a std::string into a range (v3) of std::string

How to split a std::string into a range (v3) of std::string_views?

I need to split a std::string at all spaces. The resulting range should however transform it's element to std::string_views. I'm struggling with the "element type" of the range. I guess, the type is something like a c_str. How can I transform the "split"-part into string_views?

#include <string>
#include <string_view>
#include "range/v3/all.hpp"

int main()
{
    std::string s = "this should be split into string_views";

    auto view = s 
            | ranges::view::split(' ') 
            | ranges::view::transform(std::string_view);
}

Solution

This answer is for ranges-v3 and c++17.

For a c++20/c++23 answer using std-ranges, see Peng's answer.

(One of) the problem here is that ranges::view::split returns a range of ranges, and you cannot construct a std::string_view directly from a range.

You want something like this:

auto view = s
    | ranges::views::split(' ')
    | ranges::views::transform([](auto &&rng) {
        const auto size = ranges::distance(rng);
        // &*rng.begin() is UB if the range is empty, so we need
        // to check before using it
        return size ? std::string_view(&*rng.begin(), size) : std::string_view();
});

There might be a better/easier way to do this but:

&*rng.begin() will give you the address of the first character of the chunk in the original string.
ranges::distance(rng) will give you the number of characters in this chunk. Note that this is slower than ranges::size but required here because we cannot retrieve the size of rng in constant time.