std::ranges::move_backward, std::ranges::move_backward_result - cppreference.com (original) (raw)

1. Moves the elements in the range, defined by [first, last), to another range [d_last - N, d_last), where N = ranges::distance(first, last). The elements are moved in reverse order (the last element is moved first), but their relative order is preserved. The behavior is undefined if d_last is within **(**first, last**]**. In such a case, ranges::move may 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 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 *(d_last - n) = ranges::iter_move(last - n) for each integer n, where 0 ≤ n < N.

The function-like entities described on this page are algorithm function objects (informally known as 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.

Contents

• 1 Parameters
• 2 Return value
• 3 Complexity
• 4 Notes
• 5 Possible implementation
• 6 Example
• 7 See also

[edit] Parameters

[edit] Return value

{last, d_last - N}.

[edit] Complexity

1. Exactly N 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 S1, std::bidirectional_iterator I2> requires std::indirectly_movable<I1, I2> constexpr ranges::move_backward_result<I1, I2> operator()(I1 first, S1 last, I2 d_last) const { auto i {last}; for (; i != first; *--d_last = ranges::iter_move(--i)) {} return {std::move(last), std::move(d_last)}; }   template<ranges::bidirectional_range R, std::bidirectional_iterator I> requires std::indirectly_movable<ranges::iterator_t, I> constexpr ranges::move_backward_result<ranges::borrowed_iterator_t, I> operator()(R&& r, I d_last) const { return (*this)(ranges::begin(r), ranges::end(r), std::move(d_last)); } };   inline constexpr move_backward_fn move_backward {};

[edit] Example

#include #include #include #include #include   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 : std::string{"·"}) << ' '; 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