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std::ranges::borrowed_range, std::ranges::enable_borrowed_range

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< cpp‎ | ranges
 
 
Ranges library
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(C++23)
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borrowed_range
  
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Defined in header <ranges>
template<class R>

concept borrowed_range =
    ranges::range<R> &&
    (std::is_lvalue_reference_v<R> ||

     ranges::enable_borrowed_range<std::remove_cvref_t<R>>);
(1) (since C++20)
template<class R>
inline constexpr bool enable_borrowed_range = false;
(2) (since C++20)
1) The concept borrowed_range defines the requirements of a range such that a function can take it by value and return iterators obtained from it without danger of dangling.
2) The enable_borrowed_range variable template is used to indicate whether a range is a borrowed_range. The primary template is defined as false.

Contents

[edit] Semantic requirements

Let U be std::remove_reference_t<T> if T is an rvalue reference type, and T otherwise. Given a variable u of type U, T models borrowed_range only if the validity of iterators obtained from u is not tied to the lifetime of that variable.

[edit] Specializations

Specializations of enable_borrowed_range for all specializations of the following standard templates are defined as true:

Specialization of enable_borrowed_range for the following standard range adaptors are defined as true if and only if std::ranges::enable_borrowed_range<V> is true, where V is the underlying view type:

(since C++23)

Specialization for std::ranges::zip_view is defined as true if and only if (std::ranges::enable_borrowed_range<Vs> && ...) is true, where Vs... are all view types it adapts.

(since C++23)

Users may specialize enable_borrowed_range to true for cv-unqualified program-defined types which model borrowed_range, and false for types which do not. Such specializations shall be usable in constant expressions and have type const bool.

[edit] Example

Demonstrates the specializations of enable_borrowed_range for program defined types. Such specializations protect against potentially dangling results.

#include <algorithm>
#include <array>
#include <cstddef>
#include <iostream>
#include <ranges>
#include <span>
#include <type_traits>
 
template <typename T, std::size_t N>
struct MyRange : std::array<T, N> { };
 
template <typename T, std::size_t N>
inline constexpr bool std::ranges::enable_borrowed_range<MyRange<T, N>> = false;
 
template <typename T, std::size_t N>
struct MyBorrowedRange : std::span<T, N> { };
 
template <typename T, std::size_t N>
inline constexpr bool std::ranges::enable_borrowed_range<MyBorrowedRange<T, N>> = true;
 
int main()
{
    static_assert(std::ranges::range<MyRange<int, 8>>);
    static_assert(std::ranges::borrowed_range<MyRange<int, 8>> == false);
    static_assert(std::ranges::range<MyBorrowedRange<int, 8>>);
    static_assert(std::ranges::borrowed_range<MyBorrowedRange<int, 8>> == true);
 
    auto getMyRangeByValue = [] { return MyRange<int, 4>{ {1, 2, 42, 3} }; };
    auto dangling_iter = std::ranges::max_element(getMyRangeByValue());
    static_assert(std::is_same_v<std::ranges::dangling, decltype(dangling_iter)>);
//  *dangling_iter; // compilation error (i.e. dangling protection works.)
 
    auto my = MyRange<int, 4>{ {1, 2, 42, 3} };
    auto valid_iter = std::ranges::max_element(my);
    std::cout << *valid_iter << ' '; // OK: 42
 
    auto getMyBorrowedRangeByValue = [] {
        static int sa[4] {1, 2, 42, 3};
        return MyBorrowedRange<int, std::size(sa)>{sa};
    };
    auto valid_iter2 = std::ranges::max_element(getMyBorrowedRangeByValue());
    std::cout << *valid_iter2 << '\n'; // OK: 42
}

Output:

42 42

[edit] See also

a placeholder type indicating that an iterator or a subrange should not be returned since it would be dangling
(class) [edit]