std::reverse_iterator<Iter>::operator*,->
From cppreference.com
reference operator*() const;
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(1) | (constexpr since C++17) |
| (2) | ||
pointer operator->() const;
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(constexpr since C++17) (until C++20) |
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constexpr pointer operator->() const
requires (std::is_pointer_v<Iter> ||
requires (const Iter i) { i.operator->(); });
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(since C++20) | |
Returns a reference or pointer to the element previous to current.
1) Equivalent to
Iter tmp = current; return *--tmp;.2) Equivalent to:
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(until C++11) |
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(since C++11) (until C++20) |
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(since C++20) |
Return value
Reference or pointer to the element previous to current.
Example
Run this code
#include <complex>
#include <iostream>
#include <iterator>
#include <list>
int main()
{
using RI0 = std::reverse_iterator<int*>;
int a[]{0, 1, 2, 3};
RI0 r0{std::rbegin(a)};
std::cout << "*r0 = " << *r0 << '\n';
*r0 = 42;
std::cout << "a[3] = " << a[3] << '\n';
using RI1 = std::reverse_iterator<std::list<std::complex<double>>::iterator>;
std::list<std::complex<double>> li{{1, 2}, {3, 4}, {5, 6}};
const RI1 r1{li.rbegin()};
std::cout << "*r1 = (" << r1->real() << ',' << r1->imag() << ")\n";
li.insert(li.end(), std::complex<double>{7, 8});
std::cout << "*r1 = (" << r1->real() << ',' << r1->imag() << ")\n";
}
Output:
*r0 = 3
a[3] = 42
*r1 = (5,6)
*r1 = (7,8)
Defect reports
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
| DR | Applied to | Behavior as published | Correct behavior |
|---|---|---|---|
| LWG 2188 | C++11 | operator-> used & to take address
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uses std::addressof instead |
See also
| accesses an element by index (public member function) |