Dynamic memory allocation - free - c

Question

I was looking over examples of dynamic memory allocation being used in c programs but I've noticed that many of them do not use free at the end. I was wondering if that's just an error on the programmer's part or if there are some instances in which you shouldn't free up pointers at the end. Here is an example I saw. If I am right and there should be a free where would be the appropriate place to put it? (https://www.geeksforgeeks.org/linked-list-set-2-inserting-a-node/)
// A complete working C program to demonstrate all insertion methods
// on Linked List
#include <stdio.h>
#include <stdlib.h>
// A linked list node
struct Node
{
int data;
struct Node *next;
};
/* Given a reference (pointer to pointer) to the head of a list and
an int, inserts a new node on the front of the list. */
void push(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head */
new_node->next = (*head_ref);
/* 4. move the head to point to the new node */
(*head_ref) = new_node;
}
/* Given a node prev_node, insert a new node after the given
prev_node */
void insertAfter(struct Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL)
{
printf("the given previous node cannot be NULL");
return;
}
/* 2. allocate new node */
struct Node* new_node =(struct Node*) malloc(sizeof(struct Node));
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. move the next of prev_node as new_node */
prev_node->next = new_node;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));
struct Node *last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be the last node, so make next of
it as NULL*/
new_node->next = NULL;
/* 4. If the Linked List is empty, then make the new node as head */
if (*head_ref == NULL)
{
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
return;
}
// This function prints contents of linked list starting from head
void printList(struct Node *node)
{
while (node != NULL)
{
printf(" %d ", node->data);
node = node->next;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
// Insert 6. So linked list becomes 6->NULL
append(&head, 6);
// Insert 7 at the beginning. So linked list becomes 7->6->NULL
push(&head, 7);
// Insert 1 at the beginning. So linked list becomes 1->7->6->NULL
push(&head, 1);
// Insert 4 at the end. So linked list becomes 1->7->6->4->NULL
append(&head, 4);
// Insert 8, after 7. So linked list becomes 1->7->8->6->4->NULL
insertAfter(head->next, 8);
printf("\n Created Linked list is: ");
printList(head);
return 0;
}

Answer 1

At the end of the program all memory is freed anyway so at one level it makes no difference.
Having said that the sample linked list really should have a deleteList function as this is meant to be an example of 'how to do it right'. And that delete should be called at the end of main. Somebody cutting and pasting this code will leak all over the place.
Its also a good habit to get into, tidy up after yourself. If you ever use valgrind you will want it to give you a clean bill of health
There are no circumstances under which it is ever 'bad' to free up the memory you used at the end of the program.