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std::reverse_iterator::operator*,->(3) | C++ Standard Libary | std::reverse_iterator::operator*,->(3) |
NAME¶
std::reverse_iterator::operator*,-> - std::reverse_iterator::operator*,->
Synopsis¶
reference operator*() const; (until C++17)
constexpr reference operator*() const; (since C++17)
pointer operator->() const; (until C++17)
constexpr pointer operator->() const; (since C++17)
(1) (until C++20)
constexpr pointer operator->() const (2)
requires (std::is_pointer_v<Iter> || (since C++20)
requires (const Iter i) { i.operator->(); });
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*());. (until C++20)
2) Equivalent to return current - 1; if Iter is a pointer type. (since
C++20)
Otherwise, equivalent to return std::prev(current).operator->();.
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¶
operator[] accesses an element by index
(public member function)
2022.07.31 | http://cppreference.com |