A simple question

[Razoa]

Read this piece of code:

/*An empty pointer*/
struct node *root_node;

root_node = malloc(sizeof(struct node));

free(root_node);

Does the entire instance of the structure which the pointer points to get freed up or does only the first element in the structure gets freed up. It may seem trivial but I have no way of actually testing this.

[itsme86]

It free()s the entire instance of the structure. Just remember, for every malloc() you need to have exactly 1 free(). Always.

[Tonto]

Do you mean something more like, if node was defined like

struct node
{
      struct node * next;

      // Other elements
      // Other elements
}

And you were actually like

root_node = malloc(sizeof(struct node));
rootnode->next = malloc(sizeof(struct node));
rootnode->next->next = malloc(sizeof(struct node));
rootnode->next->next->next = malloc(sizeof(struct node));
rootnode->next->next->next->next = NULL;

Would the whole list be freed? Or do you really mean the first element of the structure.

First case, no. You would make a routine to do so.
Second case, no. The whole structure is freed.

[Driver]

You'd do it in a loop, as you can't know how many you'll have.
When you create a new node (reserve the space using malloc()), set the next pointer to null until another node is created, then overwrite that null value with a pointer to the newly created node, and so on.

To release it, scan from the start and follow the chain. When the next value is null, you've reached the end. You'll need to use a temporary intermediate variable.

What you're after here is a linked list; have a look here for an article about them.

[Razoa]

Thanks for the help. You've answered my questions. I'm sorry I should have been more specific: I am in fact making a linked list.