std::reverse_iterator<Iter>::base
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< cpp | iterator | reverse iterator
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std::reverse_iterator
reverse_iterator::base
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(C++14)
iterator_type base() const;
(constexpr since C++17)
Returns the underlying iterator.
Contents
[edit] Return value
[edit] Notes
The base iterator refers to the element that is next (from the iterator_type perspective) to the element the reverse_iterator is currently pointing to. That is &*(this->base() - 1) == &*(*this).
[edit] 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 << '\n' << "*r_it == " << *r_it << '\n' << "*r_it.base() == " << *r_it.base() << '\n' << "*(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'; for (auto it = r_begin; it != r_end; ++it) std::cout << *it << ' '; std::cout << '\n'; }
Output:
*it == 3 *r_it == 2 *r_it.base() == 3 *(r_it.base() - 1) == 2 0 1 2 3 4 5 5 4 3 2 1 0