std::next
From cppreference.com
                    
                                        
                    
                    
                                                            
                    | Defined in header  <iterator> | ||
| template< class ForwardIt > ForwardIt next( | (since C++11) (until C++17) | |
| template< class InputIt > constexpr InputIt next( | (since C++17) | |
Return the nth successor of iterator it.
| Contents | 
[edit] Parameters
| it | - | an iterator | 
| n | - | number of elements to advance | 
| Type requirements | ||
| - ForwardItmust meet the requirements ofForwardIterator. | ||
| - InputItmust meet the requirements ofInputIterator. | ||
[edit] Return value
The nth successor of iterator it.
[edit] Possible implementation
| template<class ForwardIt> ForwardIt next(ForwardIt it, typename std::iterator_traits<ForwardIt>::difference_type n = 1) { std::advance(it, n); return it; } | 
[edit] 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.
[edit] 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
[edit] See also
| (C++11) | decrement an iterator (function) | 
| advances an iterator by given distance (function) | 


