std::ranges::reverse
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std::ranges
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Defined in header
<algorithm>
Call signature
template< std::bidirectional_iterator I, std::sentinel_for <I> S >
(1)
(since C++20)
requires std::permutable <I>
constexpr I
template< ranges::bidirectional_range R >
(2)
(since C++20)
requires std::permutable <ranges::iterator_t <R>>
constexpr ranges::borrowed_iterator_t <R>
1) Reverses the order of the elements in the range
[first, last). Behaves as if applying ranges::iter_swap to every pair of iterators first + i, last - i - 1 for each integer
i, where 0 ≤ i < (last - first) / 2.2) Same as (1), but uses r as the range, as if using ranges::begin (r) as first and ranges::end (r) as last.
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
[edit] Parameters
first, last
-
the iterator-sentinel pair defining the range of elements to reverse
r
-
the range of elements to reverse
[edit] Return value
An iterator equal to last.
[edit] Complexity
Exactly (last - first) / 2 swaps.
[edit] Notes
Implementations (e.g. MSVC STL) may enable vectorization when the iterator type models contiguous_iterator and swapping its value type calls neither non-trivial special member function nor ADL-found swap.
[edit] Possible implementation
See also implementations in libstdc++ and MSVC STL.
struct reverse_fn { template<std::bidirectional_iterator I, std::sentinel_for <I> S> requires std::permutable <I> constexpr I operator()(I first, S last) const { auto last2 {ranges::next (first, last)}; for (auto tail {last2}; !(first == tail or first == --tail); ++first) ranges::iter_swap (first, tail); return last2; } template<ranges::bidirectional_range R> requires std::permutable <ranges::iterator_t <R>> constexpr ranges::borrowed_iterator_t <R> operator()(R&& r) const { return (*this)(ranges::begin (r), ranges::end (r)); } }; inline constexpr reverse_fn reverse {};
[edit] Example
Run this code
#include <algorithm> #include <array> #include <iostream> #include <string> int main() { std::string s {"ABCDEF"}; std::cout << s << " → "; std::ranges::reverse(s.begin(), s.end()); std::cout << s << " → "; std::ranges::reverse(s); std::cout << s << " │ "; std::array a {1, 2, 3, 4, 5}; for (auto e : a) std::cout << e << ' '; std::cout << "→ "; std::ranges::reverse(a); for (auto e : a) std::cout << e << ' '; std::cout << '\n'; }
Output:
ABCDEF → FEDCBA → ABCDEF │ 1 2 3 4 5 → 5 4 3 2 1