std::ranges::rend
From cppreference.com
Defined in header <ranges>
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Defined in header <iterator>
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inline namespace /* unspecified */ { inline constexpr /* unspecified */ rend = /* unspecified */; |
(since C++20) (customization point object) |
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Call signature |
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template< class T > requires /* see below */ |
(since C++20) | |
Returns a sentinel indicating the end of a reversed range.
If T
is an array type and std::remove_all_extents_t<std::remove_reference_t<T>> is incomplete, then the call to ranges::rend
is ill-formed, no diagnostic required.
If the argument is an lvalue or ranges::enable_borrowed_range<std::remove_cv_t<T>> is true, then a call to ranges::rend
is expression-equivalent to:
- decay-copy(t.rend())(until C++23)auto(t.rend())(since C++23), if that expression is valid and its type models std::sentinel_for<decltype(ranges::rbegin(std::declval<T>()))>.
- Otherwise, decay-copy(rend(t))(until C++23)auto(rend(t))(since C++23), if
T
is a class or enumeration type, that expression is valid and its type models std::sentinel_for<decltype(ranges::rbegin(std::declval<T>()))>, where the meaning ofrend
is established as if by performing argument-dependent lookup only. - Otherwise, std::make_reverse_iterator(ranges::begin(t)) if both ranges::begin(t) and ranges::end(t) are valid expressions, have the same type, and that type models std::bidirectional_iterator.
In all other cases, a call to ranges::rend
is ill-formed, which can result in substitution failure when ranges::rend(t) appears in the immediate context of a template instantiation.
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