std::next vs std::advance in C++

std::advance and std::next are used to advance the iterator by a certain position, such that we can make the iterator point to a desired position. Although both have same purpose, but their implementation is different from each other. This makes it important for us to understand the difference between the two. In C++11, std::next() will advance by one by default, whereas std::advance() requires a distance.

1. Syntactical Difference: Syntax of std::advance and std::next is:

// Definition of std::advance()
template
void advance( InputIt& it, Distance n );

it: Iterator to be advanced
n: Distance to be advanced

// Definition of std::next()
ForwardIterator next (ForwardIterator it,
       typename iterator_traits::difference_type n = 1);

it: Iterator pointing to base position
n: Distance to be advanced from base position.

• Return type: std::advance does not return anything, whereas std::next returns an iterator after advancing n positions from the given base position.
• As in the syntax of std::next(), it will at least advance the iterator by one position, even if we do not specify the position which it has to advance as it has a default value one, whereas if we use std::advance, it has no such default argument.

1. Working
   • Argument Modification: std::advance modifies it arguments such that it points to the desired position, whereas, std::next does not modify its argument, infact it returns a new iterator pointing to the desired position. 

CPP

// C++ program to demonstrate
// std::advance vs std::next
#include <iostream>
#include <iterator>
#include <deque>
#include <algorithm>
using namespace std;
int main()
{
    // Declaring first container
    deque<int> v1 = { 1, 2, 3 };

    // Declaring second container for
    // copying values
    deque<int> v2 = { 4, 5, 6 };

    deque<int>::iterator ii;
    ii = v1.begin();
    // ii points to 1 in v1

    deque<int>::iterator iii;
    iii = std::next(ii, 2);
    // ii not modified

    // For std::advance
    // std::advance(ii, 2)
    // ii modified and now points to 3

    // Using copy()
    std::copy(ii, iii, std::back_inserter(v2));
    // v2 now contains 4 5 6 1 2

    // Displaying v1 and v2
    cout << "v1 = ";

    int i;
    for (i = 0; i < 3; ++i) {
        cout << v1[i] << " ";
    }

    cout << "\nv2 = ";
    for (i = 0; i < 5; ++i) {
        cout << v2[i] << " ";
    }

    return 0;
}

• Output:

v1 = 1 2 3
v2 = 4 5 6 1 2 

• Explanation: As can be seen, we want to make ii point to 2 spaces ahead of where it is pointing, so if we use std::advance ii will be pointing to two spaces ahead, whereas if we use std::next, then ii will not be advanced, but an iterator pointing to the new position will be returned, and will be stored in iii.

1. Pre-requisite: std::next requires the iterator passed as argument to be of type at least forward iterator, whereas std::advance does not have such restrictions, as it can work with any iterator, even with input iterator or better than it.

Let us see the differences in a tabular form -:

	std::next	std::advance
1.	It is used to return nth successor of an iterator	It does not have any return type.
2.	It takes two parameters that are -: number of elements and a iterator.	It takes two parameters number of elements and iterator
3.	Its Time complexity in best case is constant	Its Time complexity in best case is constant
4.	Its Time complexity in worst case is linear	Its Time complexity in worst case is linear