std::reverse_iterator::operator*,-> - cppreference.com (original) (raw)
| reference operator*() const; | (1) | (constexpr since C++17) |
|---|---|---|
| (2) | ||
| pointer operator->() const; | (constexpr since C++17) (until C++20) | |
| constexpr pointer operator->() const requires (std::is_pointer_v<Iter> | | requires (const Iter i) { i.operator->(); }); |
Returns a reference or pointer to the element previous to [current](../reverse%5Fiterator.html#current "cpp/iterator/reverse iterator").
| Member function | Equivalent to |
|---|---|
| operator* | Iter tmp = current; return *--tmp; |
| operator-> | 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) |
[edit] Return value
As described above.
[edit] Example
#include #include #include #include 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::iterator>; std::vector vi{0, 1, 2, 3}; RI1 r1{vi.rend() - 2}; std::cout << "*r1 = " << *r1 << '\n'; using RI2 = std::reverse_iterator<std::vector<std::complex>::iterator>; std::vector<std::complex> 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)
[edit] 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 |