# std::next

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< cpp‎ | iterator

Defined in header `<iterator>` |  |
template< class ForwardIt > ForwardIt next( ForwardIt it, typename std::iterator_traits<ForwardIt>::difference_type n = 1 ); |  | (since C++11) (until C++17)
template< class InputIt > constexpr InputIt next( InputIt it, typename std::iterator_traits<InputIt>::difference_type n = 1 ); |  | (since C++17)

Return the nth successor of iterator it.

### Parameters

it | - | an iterator
n | - | number of elements to advance
Type requirements
- `ForwardIt` must meet the requirements of LegacyForwardIterator.
- `InputIt` must meet the requirements of LegacyInputIterator.

### Return value

The nth successor of iterator it.

### Complexity

Linear.

However, if `InputIt` or `ForwardIt` additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.

### Possible implementation

    template<class ForwardIt>
    ForwardIt next(ForwardIt it,
                   typename std::iterator_traits<ForwardIt>::difference_type n = 1)
    {
        std::advance(it, n);
        return it;
    }

### Notes

Although the expression ++c.begin() often compiles, it is not guaranteed to do so: c.begin() is an rvalue expression, and there is no LegacyBidirectionalIterator requirement that specifies that increment of an rvalue is guaranteed to work. In particular, when iterators are implemented as pointers, ++c.begin() does not compile, while std::next(c.begin()) does.

### Example

Run this code

    #include <iostream>
    #include <iterator>
    #include <vector>

    int main()
    {
        std::vector<int> v{ 3, 1, 4 };

        auto it = v.begin();

        auto nx = std::next(it, 2);

        std::cout << *it << ' ' << *nx << '\n';
    }

Output:

    3 4

### See also

prev (C++11) | decrement an iterator (function template)
advance | advances an iterator by given distance (function template)

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