constexpr
Return the nth successor (or -nth predecessor if n is negative) of iterator it.
[edit] Parameters it - an iterator n - number of elements to advance Type requirements -InputIt
must meet the requirements of LegacyInputIterator. [edit] Return value
An iterator of type InputIt
that holds the nth successor (or -nth predecessor if n is negative) of iterator it.
Linear.
However, if InputIt
additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
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; }[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 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.
#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:
[edit] Defect reportsThe following behavior-changing defect reports were applied retroactively to previously published C++ standards.
[edit] See also decrement an iteratorRetroSearch is an open source project built by @garambo | Open a GitHub Issue
Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo
HTML:
3.2
| Encoding:
UTF-8
| Version:
0.7.4