# std::reverse_iterator<Iter>::base

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< cpp‎ | iterator‎ | reverse iterator

iterator_type base() const; |  | (constexpr since C++17)

Returns the underlying iterator.

### Return value

`current`

### 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).

### 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

### See also

operator*operator-> | accesses the pointed-to element (public member function)

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