std::reverse_iterator<Iter>::base
From cppreference.com
< cpp | iterator | reverse iterator
iterator_type base() const; |
(until C++17) | |
constexpr iterator_type base() const; |
(since C++17) | |
Returns the underlying base iterator. That is std::reverse_iterator(it).base() == it.
The base iterator refers to the element that is next (from the std::reverse_iterator::iterator_type perspective) to the element the reverse_iterator
is currently pointing to. That is &*(rit.base() - 1) == &*rit.
Parameters
(none)
Return value
The underlying iterator.
Exceptions
(none)
Example
Run this code
#include <iostream> #include <iterator> #include <vector> int main() { std::vector<int> v = { 0, 1, 2, 3, 4, 5 }; using RevIt = std::reverse_iterator<std::vector<int>::iterator>; { const auto it = v.begin() + 3; RevIt r_it(it); std::cout << "*it == " << *it << ", *r_it.base() == " << *r_it.base() << '\n' << "*r_it == " << *r_it <<", *(r_it.base()-1) == " << *(r_it.base()-1) << "\n"; } { RevIt r_end(v.begin()); RevIt r_begin(v.end()); for (auto it = r_end.base(); it != r_begin.base(); ++it) { std::cout << *it << " "; } std::cout << "\n"; } }
Output:
*it == 3, *r_it.base() == 3 *r_it == 2, *(r_it.base()-1) == 2 0 1 2 3 4 5
See also
accesses the pointed-to element (public member function) |