Defined in header <iterator>
template< class InputIt >
InputIt next( (since C++11)
InputIt it, (until C++17)
typename std::iterator_traits<InputIt>::difference_type n = 1 );
template< class InputIt >
constexpr InputIt next( (since C++17)
InputIt it,
typename std::iterator_traits<InputIt>::difference_type n = 1 );
Return the nth successor of iterator it.
it - an iterator
n - number of elements to advance
-
InputIt must meet the requirements of LegacyInputIterator.
The nth successor 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;
}
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.
// Run this code
#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
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
prev decrement an iterator
(C++11) (function template)
advance advances an iterator by given distance
(function template)
distance returns the distance between two iterators
(function template)
ranges::next increment an iterator by a given distance or to a bound
(C++20) (niebloid)