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std::reverse_iterator<Iter>::operator*,-> - cppreference.com

From cppreference.com

reference operator*() const;
(1) (constexpr since C++17) 
pointer operator->() const;
(2) (until C++20)
(constexpr since C++17)		 
constexpr pointer operator->() const
    requires (std::is_pointer_v<Iter> ||
              requires (const Iter i) { i.operator->(); });
(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:

return &(operator*());

(until C++11)

return std::addressof(operator*());

(since C++11)
(until C++20)
  • return current - 1; if Iter is a pointer type.
  • return std::prev(current).operator->();} otherwise.
 (since C++20)

Return value

Reference or pointer to the element previous to current.

Example

#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 uses std::addressof instead

See also