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

02 Aug, 2017
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.
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.WorkingArgument 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.// 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.Pre-requisite: std::next requires the iterator passed as argument to be of type at least forward iterator, whereas std::advance doesnot have such restrictions, as it can work with any iterator, even with input iterator or better than it.
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.
// 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.
WorkingArgument 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.// 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.
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.// 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.
// 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.
Pre-requisite: std::next requires the iterator passed as argument to be of type at least forward iterator, whereas std::advance doesnot have such restrictions, as it can work with any iterator, even with input iterator or better than it.
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