std::next

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

template< class InputIt >
constexpr InputIt next(
  InputIt it, 
  typename std::iterator_traits<InputIt>::difference_type n = 1 );

Return the nth successor of iterator it.

Parameters

Return value

The nth successor of iterator it.

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 BidirectionalIterator 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

#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';
}

3 4

See also