merge sort with c++20 ranges library

sort_range.cxx

#include <ranges>
#include <type_traits>
#include <vector>
#include <list>
#include <iostream>

template <typename T> requires std::ranges::sized_range<T>
T &&merge_sort(T &&arr) {
    if (std::ranges::size(arr) < 2) return std::forward<T>(arr);
    auto mid_len = std::ranges::size(arr) / 2;
    auto mid = std::ranges::next(std::ranges::begin(arr), mid_len);
    auto left = merge_sort(std::ranges::subrange(std::ranges::begin(arr), mid, mid_len));
    auto right = merge_sort(std::ranges::subrange(mid, std::ranges::end(arr), std::ranges::size(arr) - mid_len));
    auto l = left.begin(), r = right.begin();
    std::vector<std::ranges::range_value_t<T>> ret;
    ret.reserve(arr.size());
    while (l != left.end() && r != right.end()) {
        if (*l < *r) {
            ret.emplace_back(*l++);
        } else {
            ret.emplace_back(*r++);
        }
    }
    ret.insert(ret.end(), l, left.end());
    ret.insert(ret.end(), r, right.end());

    using std::copy;
    copy(ret.begin(), ret.end(), arr.begin());
    return std::forward<T>(arr);
}

int main() {
    std::list<int> arr{1, 3, 2, 8, 30, 4, 5};
    merge_sort(arr);
    for (auto &v : arr) {
        std::cout << v << std::endl;
    }

    std::cout << "-------\n";

    std::vector<int> arr1{1, 3, 2, 8, 30, 4, 5};
    merge_sort(arr1);
    for (auto &v : arr1) {
        std::cout << v << std::endl;
    }
}

Isn't it copies when passing an l value into r value ref?

By leveraging the template system and reference collapsing, the rvalue ref will collapse to lvalue ref when passing lvalue ref.