Clarification on passing a pointer by reference

Question

This is kind of silly, but I can't really explain why this is happening. As an exercise, I wanted to reverse a singly-linkedlist and I did this by defining the method:

class solution {
    void reverseLinkedList(Node*& head) {
      Node* curr = head;
      Node* prev = NULL;
      while (curr != NULL) {
        Node* _next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = _next;
      }
      head = prev;
}

In my main function, I make the call

solution s;
s.reverseLinkedList(head);

Node* iterator = head;
while (iterator != NULL) {
    std::cout<<iterator->data<<std::endl;
    iterator = iterator->next;
}

Where I previously defined my head pointer to some linkedlist. The while loop is for printing my linkedlist and the function does it job. This only worked after I passed the head node by reference; I initially tried to pass Node* head instead of Node*& head in the beginning, and it only printed the first element of my linkedlist (and without reversing it). For example, if I didn't pass by reference for a list 1->2->3, I would print out just 1.

I thought passing a pointer would be enough? Why did I get such weird behaviour without passing by reference>

Answer 1

Local variables in C++ (stored in the stack) have block scope, i.e., they run out of scope after the block in which they are defined is executed.

When you are passing in a pointer to the function, a copy of the pointer is created and that copy is what is passed. Once the function is executed, the variables in the function workspace run out of scope. Any non-static Automatic variables created within the function are destroyed.

When you pass in by reference you don't pass in a copy of the variable but you pass in the actual variable, thereby any changes made to the variable are reflected on the actual variable passed into the function(by reference).

I would like to point out that the pointer to the next node is stored in memory and has an address to the location it is stored. So if you want to not pass in by reference you can do this:

1. Use a pointer to the pointer, which points to the memory location at which pointer variable(address) to the next node is stored
2. Pass in this to the function (not by reference)
3. Dereference the pointer and store the new address you want to point to.

I know this is a bit confusing, but look into this small piece of code that adds a node to a linked list.

void addNode(Node** head, int newData)
{
    Node* newNode = new Node;
    newNode->data = newData; // Can also be done using (*newNode).data
    newNode->next = *head;
    *head = newNode;
}