std::next
From cppreference.com
                    
                                        
                    
                    
                                                            
                    |   Defined in header  <iterator>
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|   template< class ForwardIt > ForwardIt next(  | 
 (since C++11)  (until C++17)  | 
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|   template< class InputIt > constexpr InputIt next(  | 
(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.
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 -InputIt must meet the requirements of LegacyInputIterator.
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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
|    (C++11)  | 
   decrement an iterator   (function template)  | 
|    advances an iterator by given distance   (function template)  |