std::next
Aus cppreference.com
<metanoindex/>
<tbody> </tbody>| definiert in Header <iterator>
|
||
template< class ForwardIt > ForwardIt next( ForwardIt it, typename std::iterator_traits<ForwardIt>::difference_type n = 1 ); |
(seit C++11) | |
Bringen Sie den
nth Nachfolger von Iterator it .Original:
Return the
nth successor of iterator it.The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Parameter
| it | - | ein iterater '
Original: an iterater' The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| n | - | Anzahl der Elemente, um fortzufahren
Original: number of elements to advance The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| Type requirements | ||
-ForwardIt must meet the requirements of ForwardIterator.
| ||
Rückgabewert
Die
nth Nachfolger von Iterator it .Original:
The
nth successor of iterator it.The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
Mögliche Implementierung
template<class ForwardIt>
ForwardIt next(ForwardIt it, typename std::iterator_traits<ForwardIt>::difference_type n = 1)
{
std::advance(it, n);
return it;
}
|
Beispiel
#include <iostream>
#include <iterator>
#include <vector>
int main()
{
std::vector<int> v{ 3, 1, 4 };
auto it = v.begin();
auto nx = std::next(it, 2);
std::cout << *it << ' ' << *nx << '\n';
}
Output:
3 4