Namespaces
Variants
Views
Actions

std::next

From cppreference.com
< cpp‎ | iterator
 
 
Iterator library
Iterator concepts
Iterator primitives
Algorithm concepts and utilities
Indirect callable concepts
Common algorithm requirements
(C++20)
(C++20)
(C++20)
Utilities
(C++20)

Iterator adaptors
Iterator customization points
Iterator operations
(C++11)    
next
(C++11)
Range access
(C++11)(C++14)
(C++14)(C++14)    
(C++11)(C++14)
(C++14)(C++14)    
(C++17)(C++20)
(C++17)
(C++17)
 
Defined in header <iterator>
template< class InputIt >
InputIt next( InputIt it, typename std::iterator_traits<InputIt>::difference_type n = 1 );
(since C++11)
(until C++17)
template< class InputIt >

constexpr

InputIt next( InputIt it, typename std::iterator_traits<InputIt>::difference_type n = 1 );
(since C++17)

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

Contents

[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.

[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';
}

Output:

4 6
 5

[edit] 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

[edit] See also

(C++11)
decrement an iterator
(function template) [edit]
advances an iterator by given distance
(function template) [edit]
returns the distance between two iterators
(function template) [edit]
increment an iterator by a given distance or to a bound
(niebloid)[edit]