std::next
Defined in header <iterator>
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template< class InputIt > InputIt next( |
(since C++11) (until C++17) |
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template< class InputIt > constexpr InputIt next( |
(since C++17) | |
Return the n
th successor of iterator it
.
Parameters
it | - | an iterator |
n | - | number of elements to advance |
Type requirements | ||
-InputIt must meet the requirements of LegacyInputIterator.
|
Return value
The n
th successor of iterator it
.
Complexity
Linear.
However, if InputIt
additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Possible implementation
template<class InputIt> constexpr // since C++17 InputIt next(InputIt it, typename std::iterator_traits<InputIt>::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 LegacyInputIterator requirement that specifies that increment of an rvalue is guaranteed to work. In particular, when iterators are implemented as pointers or its operator++
is lvalue-ref-qualified, ++c.begin() does not compile, while std::next(c.begin()) does.
Example
#include <iostream> #include <iterator> #include <vector> int main() { std::vector<int> v{ 4, 5, 6 }; auto it = v.begin(); auto nx = std::next(it, 2); std::cout << *it << ' ' << *nx << '\n'; it = v.end(); nx = std::next(it, -2); std::cout << ' ' << *nx << '\n'; }
Output:
4 6 5
Defect reports
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
LWG 2353 | C++11 | next required LegacyForwardIterator
|
LegacyInputIterator allowed |
See also
(C++11) |
decrement an iterator (function template) |
advances an iterator by given distance (function template) | |
returns the distance between two iterators (function template) | |
(C++20) |
increment an iterator by a given distance or to a bound (niebloid) |