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std::next in C++
  • Difficulty Level : Hard
  • Last Updated : 02 Aug, 2017

std::next returns an iterator pointing to the element after being advanced by certain no. of positions. It is defined inside the header file .

It does not modify its arguments and returns a copy of the argument advanced by the specified amount. If it is a random-access iterator, the function uses just once operator + or operator – for advancing. Otherwise, the function uses repeatedly the increase or decrease operator (operator ++ or operator –) on the copied iterator until n elements have been advanced.

Syntax:

ForwardIterator next (ForwardIterator it,
       typename iterator_traits::difference_type n = 1);
it: Iterator to the base position.
difference_type: It is the numerical type that represents 
distances between iterators of the ForwardIterator type.
n: Total no. of positions by which the
iterator has to be advanced. In the syntax, n is assigned
a default value 1 so it will atleast advance by 1 position.

Returns: It returns an iterator to the element 
n positions away from it.




// C++ program to demonstrate std::next
#include <iostream>
#include <iterator>
#include <deque>
#include <algorithm>
using namespace std;
int main()
{
    // Declaring first container
    deque<int> v1 = { 1, 2, 3, 4, 5, 6, 7 };
  
    // Declaring another container
    deque<int> v2 = { 8, 9, 10 };
  
    // Declaring an iterator
    deque<int>::iterator i1;
  
    // i1 points to 1
    i1 = v1.begin();
  
    // Declaring another iterator to store return
    // value and using std::next
    deque<int>::iterator i2;
    i2 = std::next(i1, 4);
  
    // Using std::copy
    std::copy(i1, i2, std::back_inserter(v2));
    // Remember, i1 stills points to 1
    // and i2 points to 5
    // v2 now contains 8 9 10 1 2 3 4
  
    // Displaying v1 and v2
    cout << "v1 = ";
  
    int i;
    for (i = 0; i < 7; ++i) {
        cout << v1[i] << " ";
    }
  
    cout << "\nv2 = ";
    for (i = 0; i < 7; ++i) {
        cout << v2[i] << " ";
    }
  
    return 0;
}


Output:

v1 = 1 2 3 4 5 6 7
v2 = 8 9 10 1 2 3 4

How can it be helpful ?



  • Advancing iterator in Lists: Since, lists support bidirectional iterators, which can be incremented only by using ++ and – – operator. So, if we want to advance the iterator by more than one position, then using std::next can be extremely useful.




    // C++ program to demonstrate std::next
    #include <iostream>
    #include <iterator>
    #include <list>
    #include <algorithm>
    using namespace std;
    int main()
    {
        // Declaring first container
        list<int> v1 = { 1, 2, 3, 7, 8, 9 };
      
        // Declaring second container
        list<int> v2 = { 4, 5, 6 };
      
        list<int>::iterator i1;
        i1 = v1.begin();
        // i1 points to 1 in v1
      
        list<int>::iterator i2;
        // i2 = v1.begin() + 3;
        // This cannot be used with lists
        // so use std::next for this
      
        i2 = std::next(i1, 3);
      
        // Using std::copy
        std::copy(i1, i2, std::back_inserter(v2));
        // v2 now contains 4 5 6 1 2 3
      
        // Displaying v1 and v2
        cout << "v1 = ";
      
        int i;
        for (i1 = v1.begin(); i1 != v1.end(); ++i1) {
            cout << *i1 << " ";
        }
      
        cout << "\nv2 = ";
        for (i1 = v2.begin(); i1 != v2.end(); ++i1) {
            cout << *i1 << " ";
        }
      
        return 0;
    }

    
    

    Output:

    v1 = 1 2 3 7 8 9
    v2 = 4 5 6 1 2 3  
    

    Explanation: Here, just look how if we want copy only a selected portion of the list, then we can make use of std::next, as otherwise we cannot use any +=, -= operators with bidirectional iterators supported by lists. So, we used std::next and directly advanced the iterator by three positions.

This article is contributed by Mrigendra Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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