std::next

Return the n^th successor (or -n^th predecessor if n is negative) of iterator it.

[edit] Parameters

[edit] Return value

An iterator of type InputIt that holds the n^th successor (or -n^th predecessor if n is negative) of iterator it.

[edit] Complexity

Linear.

However, if InputIt additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.

[edit] 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;
}

[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.

[edit] 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';
}

4 6
 5

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

[edit] See also