std::ranges...ward_result(3)

okay bestie, welcome to std::ranges::copy_backward and std::ranges::copy_backward_result from the C++ Standard Libary manual (yes, that's how it's spelled on the page 😭, section 3, page std::ranges...ward_result(3)). This is the whole glow-up, every detail kept, no cap. 💅✨

🌸 NAME 🌸

std::ranges::copy_backward,std::ranges::copy_backward_result — std::ranges::copy_backward,std::ranges::copy_backward_result

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✨ Synopsis ✨

Defined in header <algorithm>

Call signature:

template< std::bidirectional_iterator I1, std::sentinel_for<I1> S1,
          std::bidirectional_iterator I2 >
    requires std::indirectly_copyable<I1, I2>
constexpr copy_backward_result<I1, I2>
    copy_backward( I1 first, S1 last, I2 result );

(1) (since C++20)

template< ranges::bidirectional_range R, std::bidirectional_iterator I >
    requires std::indirectly_copyable<ranges::iterator_t<R>, I>
constexpr copy_backward_result<ranges::borrowed_iterator_t<R>, I>
    copy_backward( R&& r, I result );

(2) (since C++20)

Helper types:

template< class I1, class I2 >
using copy_backward_result = ranges::in_out_result<I1, I2>;

(3) (since C++20)

1. Copies the elements from the range, defined by [first, last), to another range [result - N, result), where N = ranges::distance(first, last). The elements are copied in reverse order (the last element is copied first), but their relative order is preserved.

⚠️ Undefined behavior alert, bestie: the behavior is undefined if result is within (first, last]. In such a case std::ranges::copy can be used instead. 💀

2. Same as (1), but uses r as the source range, as if using ranges::begin(r) as first, and ranges::end(r) as last.

The function-like entities described on this page are niebloids, that is:

• Explicit template argument lists cannot be specified when calling any of them.
• None of them are visible to argument-dependent lookup.
• When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

In practice, they may be implemented as function objects, or with special compiler extensions.

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🎀 Parameters 🎀

• first — the beginning of the range of elements to copy from
• last — the end of the range of elements to copy from
• r — the range of the elements to copy from
• result — the end of the destination range

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💫 Return value 💫

{last, result - N}

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⚡ Complexity ⚡

Exactly N assignments. periodt.

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🌼 Notes 🌼

When copying overlapping ranges, ranges::copy is appropriate when copying to the left (beginning of the destination range is outside the source range) while ranges::copy_backward is appropriate when copying to the right (end of the destination range is outside the source range). iykyk 🦋

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🧚 Possible implementation 🧚

Here's the vibe of how it might be implemented — exact code, no edits, bestie:

struct copy_backward_fn
{
    template<std::bidirectional_iterator I1, std::sentinel_for<I1> S1,
             std::bidirectional_iterator I2>
    requires std::indirectly_copyable<I1, I2>
    constexpr ranges::copy_backward_result<I1, I2>
        operator()(I1 first, S1 last, I2 result) const
    {
        I1 last1 {ranges::next(first, std::move(last))};
        for (I1 i {last1}; i != first;)
            *--result = *--i;
        return {std::move(last1), std::move(result)};
    }

    template<ranges::bidirectional_range R, std::bidirectional_iterator I>
    requires std::indirectly_copyable<ranges::iterator_t<R>, I>
    constexpr ranges::copy_backward_result<ranges::borrowed_iterator_t<R>, I>
        operator()(R&& r, I result) const
    {
        return (*this)(ranges::begin(r), ranges::end(r), std::move(result));
    }
};

inline constexpr copy_backward_fn copy_backward{};

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🌈 Example 🌈

// Run this code

#include <algorithm>
#include <iostream>
#include <ranges>
#include <string_view>
#include <vector>

void print(std::string_view rem, std::ranges::forward_range auto const& r)
{
    for (std::cout << rem << ": "; auto const& elem : r)
        std::cout << elem << ' ';
    std::cout << '\n';
}

int main()
{
    const auto src = {1, 2, 3, 4};
    print("src", src);

    std::vector<int> dst(src.size() + 2);
    std::ranges::copy_backward(src, dst.end());
    print("dst", dst);

    std::ranges::fill(dst, 0);
    const auto [in, out] =
        std::ranges::copy_backward(src.begin(), src.end() - 2, dst.end());
    print("dst", dst);

    std::cout
        << "(in - src.begin) == " << std::distance(src.begin(), in) << '\n'
        << "(out - dst.begin) == " << std::distance(dst.begin(), out) << '\n';
}

Output

src: 1 2 3 4
dst: 0 0 1 2 3 4
dst: 0 0 0 0 1 2
(in - src.begin) == 2
(out - dst.begin) == 4

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🔥 See also 🔥

• ranges::copy (C++20) and ranges::copy_if (C++20) — copies a range of elements to a new location (niebloid)
• ranges::copy_n (C++20) — copies a number of elements to a new location (niebloid)
• ranges::remove_copy (C++20) and ranges::remove_copy_if (C++20) — copies a range of elements omitting those that satisfy specific criteria (niebloid)
• ranges::replace_copy (C++20) and ranges::replace_copy_if (C++20) — copies a range, replacing elements satisfying specific criteria with another value (niebloid)
• ranges::reverse_copy (C++20) — creates a copy of a range that is reversed (niebloid)
• ranges::rotate_copy (C++20) — copies and rotate a range of elements (niebloid)
• ranges::unique_copy (C++20) — creates a copy of some range of elements that contains no consecutive duplicates (niebloid)
• ranges::move (C++20) — moves a range of elements to a new location (niebloid)
• ranges::move_backward (C++20) — moves a range of elements to a new location in backwards order (niebloid)
• copy_backward — copies a range of elements in backwards order (function template)

That's the whole page, bestie — source http://cppreference.com, version/date 2024.06.10, page std::ranges...ward_result(3). Stay iconic. 🫶💖✨