std::reverse_iterator<Iter>::operator*,->
From cppreference.com
< cpp | iterator | reverse iterator
(1) | ||
reference operator*() const; |
(until C++17) | |
constexpr reference operator*() const; |
(since C++17) | |
(2) | ||
pointer operator->() const; |
(until C++17) | |
constexpr pointer operator->() const; |
(since C++17) (until C++20) |
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constexpr pointer operator->() const requires (std::is_pointer_v<Iter> || |
(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 std::addressof(operator*());.
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(until C++20) |
2) Equivalent to return current - 1; if Iter is a pointer type. Otherwise, equivalent to return std::prev(current).operator->();.
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(since C++20) |
Parameters
(none)
Return value
Reference or pointer to the element previous to current
.
Example
Run this code
#include <complex> #include <iostream> #include <iterator> #include <vector> 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::vector<int>::iterator>; std::vector<int> vi { 0, 1, 2, 3 }; RI1 r1 { vi.rend() - 2 }; std::cout << "*r1 = " << *r1 << '\n'; using RI2 = std::reverse_iterator<std::vector<std::complex<double>>::iterator>; std::vector<std::complex<double>> vc { {1,2}, {3,4}, {5,6}, {7,8} }; RI2 r2 { vc.rbegin() + 1 }; std::cout << "vc[2] = " << "(" << r2->real() << "," << r2->imag() << ")\n"; }
Output:
*r0 = 3 a[3] = 42 *r1 = 1 vc[2] = (5,6)
See also
accesses an element by index (public member function) |