Question

I can’t understand where’s the problem with this very simple list I’ve been testing. The idea is to get the item at position i in a list.
I know usually I wouldn’t use a list to do that.
However, this works when I set item = 11, item = 12 and item = 13 (output would be, respectively, at position {1, 2, 3} there's the item {11, 12, 13}), BUT it does not work when I set item = 10, as the output is at position 0 there's the item 6.

int main(void)
{
    list<int> L;
    L.push_back(10);
    L.push_back(11);
    L.push_back(12);
    L.push_back(13);

    int item = 10;
    int pos;

    list<int>::iterator it = moveToItem(item, L, pos);

    cout << "at position " << pos << " there's the item " << *it;
}

list<int>::iterator moveToItem(int item, list<int> L, int& pos)
{
     pos = 0;
     list<int>::iterator it = L.begin();

     while(*it != item)
     {
         it++;
         pos++;
     }
     return it;
}

Answer 1

A copy of the list L is being made when moveToItem() is called so the returned iterator is referring to an item of a list that has been destructed. Pass the list by reference instead:

list<int>::iterator moveToItem(int item, list<int>& L, int& pos)
                                                //^

You should also protect against going past the end() of the list in the while condition, before dereferencing it.

If this is not an exercise consider using the STL algorithm std::find() and std::distance() instead:

#include <iterator>
#include <algorithm>

std::list<int>::iterator it = std::find(L.begin(), L.end(), 41);
if (it != L.end())
{
    std::cout << "at position "
              << std::distance(L.begin(), it)
              << " there's the item "
              << *it
              << "\n";
}