How to move or delete a specific element in std::forward_list?

Question

I've been trying to work this out for more than an hour. I need to move an integer from middle of the list to the beginning. There is no option for moving an integer, it would be fine if I could insert one at the beginning and delete the old one from middle, but you can't do that either. (you can delete every entry of a specific integer, not a single one)

I read everything there was about std::forward_list on cplusplus.com and googled this for 15 minutes, no results.

Answer 1

You can use Dobly Link list. ( i hope simple link list can also do your task) Here is the link for doubly link list. https://www.hackerearth.com/practice/notes/doubly-linked-list-data-structure-in-c/ Follow these steps: 1. Copy middle element. 2. add that copied element to the start of list. 3. remove the middle element. If you are using some library of list that is restricting you, make your own class of list of which removal from any index is possible.

Answer 2

Combine these three points to erase a given element (given position) in forward_list :

• Use erase_after that remove the one (or range) after the a position.

• Use two iterators, prev and it, where prev keep the previous position and it keep the actual position.

• Use before_begin to initialize prev and begin to initialize it. This is very important to be able to delete the element at the first position if it matches.

  std::forward_list<int> mylist = {10, 10, 20, 30, 40, 10, 50, 10, 60};
  
  int skey = 10;
  
  for (int & val : mylist) std::cout<<val<<", ";
  
  auto prev = mylist.before_begin();
  for (auto it = mylist.begin(); it!=mylist.end(); ++it)
  {
      if(*it==skey) // or if(it== input_it_pos) for a known position
      {
          mylist.erase_after(prev);
          break;
      }
  
      prev=it;
  }
  
  std::cout<<"\n after deletion : "<<std::endl;
  for (int & val : mylist) std::cout<<val<<", ";
  

Note 1: you can change the code to delete all the elements that match in following manner:

auto prev = mylist.before_begin();
for (auto it = mylist.begin(); it!=mylist.end(); )
{
    if(*it==skey)
    {
        it = mylist.erase_after(prev);
        // break; // Comment or uncomment to deal with only the first or all the found element(s).
    }
    else
    {
        prev = it;
        ++it;
    }
}

Note 2: if you don't use before_begin, and the two prev and it begins from the same point: (forward_list.begin()) , you can check the first element alone and then use pop_front to delete it.

you can use remove, remove_if to remove all the elements from your forward_list (remove: that compare equal to val, remove_if: which Predicate pred returns true).

Answer 3

Here you are!

// Example program
#include <iostream>
#include <string>
#include <forward_list>

void EraseAtPosition(std::forward_list<int>::iterator pos, std::forward_list<int>& your_list) {
    // Find the position before input pos
    std::forward_list<int>::iterator prev;
    for (auto it = your_list.before_begin(); it != your_list.end();) {
        prev = it;
        if (++it == pos) {
            break;
        }
    }

    // Erase out of list
    your_list.erase_after(prev);
}

int main()
{
    std::forward_list<int> l = {1, 2, 3, 5};
    std::forward_list<int>::iterator target = l.begin();
    EraseAtPosition(target, l);

    for (auto it = l.begin(); it != l.end(); ++it) {
        std::cout << *it << "\n";
    }
    return 0;
}