std::reverse_iterator<Iter>::base (3) - Linux Manuals

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

NAME

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