Circular linked list crashes when displayed - c

Question

I'm trying to make a circular linked list. When I try to display the list after creating it, the program keeps on crashing. Here is my code:
#include <stdio.h>
#include <stdlib.h>
typedef struct node {
int data;
struct node * next;
} node;
node * createList(int);
void display(node * head);
int main() {
struct node * head;
head = createList(5);
display(head);
}
node * createList(int n) {
int i = 0,data = 0;
struct node * head = NULL;
struct node * temp = NULL;
struct node * p = NULL;
for (i = 0; i < n; i++) {
temp = (node*)malloc(sizeof(node));
temp->data = data++;
temp->next = head;
if (head == NULL) {
head = temp;
} else {
p = head;
while (p->next != NULL) {
p = p->next;
}
p->next = temp;
}
}
return head;
}
void display(node * head) {
struct node * temp = head->next;
while (temp != head) {
printf("%d-> \t",temp->data);
temp = temp->next;
}
printf("\n");
}
What am I doing wrong?

Answer 1

You have set every temp's next to head in temp->next = head; but did it too early (the first is just NULL). Then you tested p->next against NULL in while (p->next != NULL) { but you should have tested against head. Alternatively, you can continue to test against NULL but then you need to initialize temp->next to NULL and assign head to temp->next only after the for loop.
Your display code started from the second link.
Here is a fixed code using the first option in 1. above:
for (i = 0; i < n; i++) {
temp = (node*)malloc(sizeof(node));
temp->data = data++;
if (head == NULL) {
head = temp;
} else {
p = head;
while (p->next != head) {
p = p->next;
}
p->next = temp;
}
temp->next = head;
}
Here is a fixed code using the alternative option in 1. above. You still need to initialize temp->next to NULL since malloc() does not initialize.
for (i = 0; i < n; i++) {
temp = (node*)malloc(sizeof(node));
temp->data = data++;
temp->next = NULL;
if (head == NULL) {
head = temp;
} else {
p = head;
while (p->next != NULL) {
p = p->next;
}
p->next = temp;
}
}
if (temp != NULL) {
temp->next = head;
}
But in reality, there is no need to "walk" from the head on every creation. You can simply keep the previous and link it to the next:
for (i = 0; i < n; i++) {
temp = (node*)malloc(sizeof(node));
temp->data = data++;
if (head == NULL) {
head = p = temp;
} else {
p = p->next = temp;
}
}
if (temp != NULL) {
temp->next = head;
}
Here is a fix for the display():
void display(node * head) {
struct node * temp = head;
if (temp != NULL) {
do {
printf("%d-> \t",temp->data);
temp = temp->next;
} while (temp != head);
}
printf("\n");
}

Answer 2

The problem is on the first node you initialize:
struct node *head = NULL;
...
for (i = 0; i < n; i++) {
...
temp->next = head;
So tmp->next == NULL on the first iteration leaving head->next == NULL. That will not work for a circular list. When you attempt to insert the 2nd node:
p = head;
while (p->next != NULL) {
What was head->next again?? (oh, NULL) Dereferencing a NULL pointer (BOOM Segfault!)
Do your circular list correctly. On insertion of the first node set:
if (head == NULL) {
head = temp;
head->next = temp; /* you must set head->next to temp */
} ...
So on the insertion of the remaining nodes you simply need:
} else {
p = head;
while (p->next != head) { /* iterate to last node */
p = p->next;
}
p->next = temp; /* now set p->next = temp */
}
Now, you handle your display() function the same way, e.g.
void display (node *head)
{
if (!head) { /* validate list not empty */
puts ("(list-empty)");
return;
}
struct node *temp = head;
do { /* same loop problem fixed in display() */
printf ("%d-> \t", temp->data);
temp = temp->next;
} while (temp != head);
putchar ('\n');
}
If you make the changes, then you can test your list with:
int main (void) {
struct node *head, *tmp;
head = createList(5);
display (head);
puts ("\niterate from mid-list");
tmp = head;
tmp = tmp->next;
tmp = tmp->next;
display (tmp);
}
Example Use/Output
$ ./bin/lls_circular_fix
0-> 1-> 2-> 3-> 4->
iterate from mid-list
2-> 3-> 4-> 0-> 1->
Lastly, you are not multiplying the type node by head in struct node * head = NULL; Write it as struct node *head = NULL; (the same for all your function declarations as well) Much more readable.
When you go to delete a note from the list, you must create a special case for both head and tail (the last node). In this sense, the singly-linked list takes a bit more effort than a doubly-linked list due to not having a prev node pointer to track the prior node.
Look things over and let me know if you have questions.
A full example would be:
#include <stdio.h>
#include <stdlib.h>
typedef struct node {
int data;
struct node *next;
} node;
node *createList (int);
void display (node *head);
int main (void) {
struct node *head, *tmp;
head = createList(5);
display (head);
puts ("\niterate from mid-list");
tmp = head;
tmp = tmp->next;
tmp = tmp->next;
display (tmp);
}
node *createList (int n)
{
int i = 0,data = 0;
struct node *head = NULL;
struct node *temp = NULL;
struct node *p = NULL;
for (i = 0; i < n; i++) {
if (!(temp = malloc (sizeof *temp))) {
perror ("malloc-temp");
return NULL;
}
temp->data = data++;
temp->next = head; /* head is NULL on 1st node insertion */
if (head == NULL) {
head = temp;
head->next = temp; /* you must set head->next to temp */
} else {
p = head;
while (p->next != head) { /* iterate to last node */
p = p->next;
}
p->next = temp; /* now set p->next = temp */
}
}
return head;
}
void display (node *head)
{
if (!head) { /* validate list not empty */
puts ("(list-empty)");
return;
}
struct node *temp = head;
do { /* same loop problem fixed in display() */
printf ("%d-> \t", temp->data);
temp = temp->next;
} while (temp != head);
putchar ('\n');
}