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