Std::reverse_iterator::base (3) Linux Manual Page

std::reverse_iterator<Iter>::base – std::reverse_iterator<Iter>::base

Synopsis

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

std::reverse_iterator::base (3) Linux Manual Page