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

  • Difficulty Level : Hard
  • Last Updated : 02 Aug, 2017
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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.
  2. 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.




      // 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.

  3. 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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