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std::ranges::move_backward, std::ranges::move_backward_result

From cppreference.com
< cpp‎ | algorithm‎ | ranges
 
 
Algorithm library
Constrained algorithms and algorithms on ranges (C++20)
Constrained algorithms: std::ranges::copy, std::ranges::sort, ...
Execution policies (C++17)
Non-modifying sequence operations
(C++11)(C++11)(C++11)
(C++17)
Modifying sequence operations
Operations on uninitialized storage
Partitioning operations
Sorting operations
(C++11)
Binary search operations
Set operations (on sorted ranges)
Heap operations
(C++11)
Minimum/maximum operations
(C++11)
(C++17)

Permutations
Numeric operations
C library
 
Constrained algorithms
Non-modifying sequence operations
Modifying sequence operations
Operations on uninitialized storage
Partitioning operations
Sorting operations
Binary search operations
Set operations (on sorted ranges)
Heap operations
Minimum/maximum operations
Permutations
 
Defined in header <algorithm>
Call signature
template<std::bidirectional_iterator I1, std::sentinel_for<I1> S1,

         std::bidirectional_iterator I2>
  requires std::indirectly_movable<I1, I2>
    constexpr ranges::move_backward_result<I1, I2>

      ranges::move_backward( I1 first, S1 last, I2 result );
(1) (since C++20)
template<ranges::bidirectional_range R, std::bidirectional_iterator I>

  requires std::indirectly_movable<ranges::iterator_t<R>, I>
    constexpr ranges::move_backward_result<ranges::borrowed_iterator_t<R>, I>

      ranges::move_backward( R&& r, I result );
(2) (since C++20)
Helper types
template<class I, class O>
  using move_backward_result = ranges::in_out_result<I, O>;
(3) (since C++20)
1) Moves the elements in the range, defined by [first, last), to another range [result - N, result), where N = last - first. The elements are moved in reverse order (the last element is moved first), but their relative order is preserved. Precondition: result is not within (first, last], otherwise the behavior is undefined and std::ranges::move should be used instead. Note: the elements in the moved-from range will still contain valid values of the appropriate type, but not necessarily the same values as before the move, as if using *(result - n) = ranges::iter_move(last - n) for each integer n, where 0 ≤ n < N.
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:

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

Contents

[edit] Parameters

first - the beginning of the range of elements to move
last - the end of the range of elements to move
r - the range of the elements to move
result - the beginning of the destination range

[edit] Return value

An object {last, result - N} .

[edit] Complexity

1) Exactly last - first move assignments.
2) Exactly std::ranges::size(r) move assignments.

[edit] Notes

When moving overlapping ranges, ranges::move is appropriate when moving to the left (beginning of the destination range is outside the source range) while ranges::move_backward is appropriate when moving to the right (end of the destination range is outside the source range).

[edit] Possible implementation

struct move_backward_fn {
  template<std::bidirectional_iterator I1, std::sentinel_for<I1> S1,
           std::bidirectional_iterator I2>
    requires std::indirectly_movable<I1, I2>
      constexpr ranges::move_backward_result<I1, I2>
        operator()( I1 first, S1 last, I2 result ) const {
          auto i {last};
          for (; i != first; *--result = ranges::iter_move(--i));
          return {std::move(last), std::move(result)};
        }
 
  template<ranges::bidirectional_range R, std::bidirectional_iterator I>
    requires std::indirectly_movable<ranges::iterator_t<R>, I>
      constexpr ranges::move_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 move_backward_fn move_backward{};

[edit] Example

#include <algorithm>
#include <iostream>
#include <string>
#include <string_view>
#include <vector>
using namespace std::literals;
 
using Vec = std::vector<std::string>;
 
void print(std::string_view rem, Vec const& vec) {
    std::cout << rem << "[" << vec.size() << "]: ";
    for (const std::string& s : vec)
        std::cout << (s.size() ? s : "·"s) << ' ';
    std::cout << '\n';
}
 
int main()
{
    Vec a{"▁", "▂", "▃", "▄", "▅", "▆", "▇", "█"};
    Vec b(a.size());
 
    print("Before move:\n" "a", a);
    print("b", b);
 
    std::ranges::move_backward(a, b.end());
 
    print("\n" "Move a >> b:\n" "a", a);
    print("b", b);
 
    std::ranges::move_backward(b.begin(), b.end(), a.end());
    print("\n" "Move b >> a:\n" "a", a);
    print("b", b);
 
    std::ranges::move_backward(a.begin(), a.begin()+3, a.end());
    print("\n" "Overlapping move a[0, 3) >> a[5, 8):\n" "a", a);
}

Possible output:

Before move:
a[8]: ▁ ▂ ▃ ▄ ▅ ▆ ▇ █
b[8]: · · · · · · · ·
 
Move a >> b:
a[8]: · · · · · · · ·
b[8]: ▁ ▂ ▃ ▄ ▅ ▆ ▇ █
 
Move b >> a:
a[8]: ▁ ▂ ▃ ▄ ▅ ▆ ▇ █
b[8]: · · · · · · · ·
 
Overlapping move a[0, 3) >> a[5, 8):
a[8]: · · · ▄ ▅ ▁ ▂ ▃

[edit] See also

moves a range of elements to a new location
(niebloid) [edit]
copies a range of elements to a new location
(niebloid) [edit]
copies a range of elements in backwards order
(niebloid) [edit]
(C++11)
moves a range of elements to a new location
(function template) [edit]
(C++11)
obtains an rvalue reference
(function template) [edit]