[simple] Deep copy of a linked list

[Gldnbr]

I'm trying to perform a deep copy of a linked list and print out the length and sum of the copied linked list, however, there is some kind of problem and I can't pinpoint it.

Can someone figure out what is wrong?

#include <iostream>
using namespace std;

struct listrec
{
	int value;
	listrec *next;
};

int listsize(listrec *top)
{
    listrec *pt; // creates pointer
    pt = top;
    int i = 0;
    while(pt != NULL)
    {
        i += 1;
        pt = pt->next;
    }
    return i;
}

int listsum(listrec *top)
{
    listrec *pt; // creates pointer
    pt = top;
    int sum = 0;
    while(pt != NULL)
    {
        sum += pt->value;
        pt = pt->next;
    }
    return sum;
}

void deepCopy(listrec* x, listrec* &copy)
{
	while (x != NULL)
	{
		{
			listrec* copy = new listrec;
			copy->value = x->value;
			copy->next = x->next;
			x = x->next;
		}
	}
}

int main()
{
    listrec *ptr, *head, *head2; // create 2 pointers (ptr, head)
    int a = 4, b = 5, c = 3; // definitions
    head = new listrec; // creates new node
    head->value = a;
    head->next = new listrec; // creates new node
    ptr = head; // sets up a linked list location
    ptr = ptr->next;
    ptr->value = b;
    ptr->next = new listrec; // creates new node
    ptr = ptr->next;
    ptr->value = c;
    ptr->next = NULL;

    cout << listsize(head) << endl; // prints the size of linked list
    cout << listsum(head) << endl; // prints the sum of all values in linked list

	deepCopy(head, head2);

	cout << listsize(head2) << endl;
	cout << listsum(head2) << endl;

    system("PAUSE");
    return 0;
}

[Peter87]

The problem with deepCopy is that it points the next pointer of the new nodes to the nodes in the old list.

[GFreak45]

I would do it like this:

listrec* deepCopy(const listrec*&copy)
{
	listrec* result = new listrec (*copy),
		*copyiterator = result;
	for (listrec* iterator = copy->next; listrec != nullptr; listrec = listrec->next)
	{
		copyiterator->next = new listrec (*iterator);
		copyiterator = copyiterator->next;
	}
	copyiterator->next = nullptr;
	return result;
}

the line: new listrec (*copy) creates an exact copy of the copy argument, however, the ->next member of this copy is still the same as the original copy member's ->next, so we need to iterate through all the items in the linked list creating copies, which is what we are doing with the:

for (listrec* iterator = copy->next; listrec != nullptr; listrec = listrec->next)
{
	copyiterator->next = new listrec (*iterator);
	copyiterator = copyiterator->next;
}

NOTE: you should probably replace nullptr with NULL in that function seeing as you are using NULL, nullptr means null pointer but its not the same as NULL