std::ranges::rend
Defined in header <ranges>
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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 */ |
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Returns a sentinel indicating the end of a reversed range.
Let t
be an object of type T
. 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:
- std::forward<T>(t).rend(), if that expression is valid, and its return type models std::sentinel_for<decltype(ranges::rbegin(std::forward<T>(t)))>.
- Otherwise, rend(std::forward<T>(t)), if
T
is a class or enumeration type, the aforementioned unqualified call is valid, and its return type models std::sentinel_for<decltype(ranges::rbegin(std::forward<T>(t)))>, where the overload resolution is performed with the following candidates:- void rend(auto&) = delete;
- void rend(const auto&) = delete;
- any declarations of
rend
found by argument-dependent lookup.
- Otherwise, std::make_reverse_iterator(ranges::begin(std::forward<T>)) if both ranges::begin(std::forward<T>(t)) and ranges::end(std::forward<T>(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.
Contents |
Expression-equivalent
Expression e is expression-equivalent to expression f, if
- e and f have the same effects, and
- either both are constant subexpressions or else neither is a constant subexpression, and
- either both are potentially-throwing or else neither is potentially-throwing (i.e. noexcept(e) == noexcept(f)).
Customization point objects
The name ranges::rend
denotes a customization point object, which is a const function object of a literal semiregular
class type. For exposition purposes, the cv-unqualified version of its type is denoted as __rend_fn
.
All instances of __rend_fn
are equal. The effects of invoking different instances of type __rend_fn
on the same arguments are equivalent, regardless of whether the expression denoting the instance is an lvalue or rvalue, and is const-qualified or not (however, a volatile-qualified instance is not required to be invocable). Thus, ranges::rend
can be copied freely and its copies can be used interchangeably.
Given a set of types Args...
, if std::declval<Args>()... meet the requirements for arguments to ranges::rend
above, __rend_fn
models
- std::invocable<__rend_fn, Args...>,
- std::invocable<const __rend_fn, Args...>,
- std::invocable<__rend_fn&, Args...>, and
- std::invocable<const __rend_fn&, Args...>.
Otherwise, no function call operator of __rend_fn
participates in overload resolution.
Notes
If the argument is an rvalue (i.e. T
is an object type) and ranges::enable_borrowed_range<std::remove_cv_t<T>> is false, or if it is of an array type of unknown bound, the call to ranges::rend
is ill-formed, which also results in substitution failure.
If ranges::rend(std::forward<T>(t)) is valid, then decltype(ranges::rend(std::forward<T>(t))) and decltype(ranges::begin(std::forward<T>(t))) model std::sentinel_for in all cases, while T
modeling std::ranges::range.
Example
#include <algorithm> #include <iostream> #include <ranges> #include <vector> int main() { std::vector<int> v = { 3, 1, 4 }; namespace ranges = std::ranges; if (ranges::find(v, 5) != ranges::rend(v)) { std::cout << "found a 5 in vector v!\n"; } int a[] = { 5, 10, 15 }; if (ranges::find(a, 5) != ranges::rend(a)) { std::cout << "found a 5 in array a!\n"; } }
Output:
found a 5 in array a!
See also
(C++20) |
returns a reverse end iterator to a read-only range (customization point object) |
(C++20) |
returns a reverse iterator to a range (customization point object) |
(C++14) |
returns a reverse end iterator for a container or array (function template) |