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.
Contents |
[edit] Parameters
it | - | an iterator |
n | - | number of elements to advance |
Type requirements | ||
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ForwardIt must meet the requirements of ForwardIterator .
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InputIt must meet the requirements of InputIterator .
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[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)
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decrement an iterator (function) |
advances an iterator by given distance (function) |