Unable to create and display linked list in C - c

Question

I want to create integer linked list and display. Suppose there are 3 nodes with values 11,22,33 . But when I display it, its printing only 1st value i.e. 11 . What is going wrong?
NOTE : To create and display linked list , Whether head and p node variable are enough or is it must to take 3 node pointer variables . i.e. head , p and q also?
#include <stdio.h>
#include<stdlib.h>
typedef struct node
{
int data;
struct node *next;
}node;
int main()
{
int i, j, num, value;
node *p = NULL;
node *head = NULL;
printf("how many nodes\r\n");
scanf("%d",&num);
for(i = 0 ;i < num ; i++)
{
printf("enter node %d = ",i+1);
scanf("%d",&value);
p = (node *)malloc(sizeof(node));
p->data = value;
p->next = NULL;
if(head == NULL)
{
head = p;
}
}
printf("linked list formed is \r\n");
for(p = head ; p != NULL ; p = p->next)
{
printf("p->data = %d\r\n ",p->data);
}
return 0;
}

Answer 1

You build a forward-chained linked list in input order by constantly updating a target point on which to hang the next node.
Initially that pointer is the head pointer.
The next node will be hung on the next pointer of that previous node.
When done, the last next pointer is set to NULL and you're finished.
It may sound complicated, but utilizing a pointer-to-pointer makes the algorithm surprisingly simple, efficient, and requires no special tedious case for testing for a null head pointer that will only ever be true once. Including added error checking
#include <stdio.h>
#include <stdlib.h>
typedef struct node
{
int data;
struct node *next;
} node;
int main()
{
node *head = NULL;
node **pp = &head;
int num;
printf("how many nodes\r\n");
if (scanf("%d", &num) == 1 && num > 0)
{
for (int i=0; i<num;)
{
int value;
printf("enter node %d = ", i+1);
if (scanf("%d", &value) == 1)
{
*pp = malloc(sizeof **pp);
if (*pp == NULL)
{
perror("Failed to allocate new list node");
exit(EXIT_FAILURE);
}
// hang the new node
(*pp)->data = value;
// setup pp to hold address of next pointer
// to populate on the next iteration.
pp = &(*pp)->next;
// next node
++i;
}
else
{
int c;
while ((c = fgetc(stdin)) != EOF && c != '\n');
}
}
// terminate the list
*pp = NULL;
}
printf("linked list formed is:\n");
for (const node *p = head; p != NULL; p = p->next)
{
printf(" p->data = %d\n", p->data);
}
// free the list
while (head)
{
node *p = head;
head = head->next;
free(p);
}
return 0;
}
Sample Run
how many nodes
5
enter node 1 = 1
enter node 2 = 3
enter node 3 = 5
enter node 4 = 7
enter node 5 = 9
linked list formed is:
p->data = 1
p->data = 3
p->data = 5
p->data = 7
p->data = 9

Answer 2

You are just updating the head first time and not creating any links
please find the fixed code below
#include <stdio.h>
#include<stdlib.h>
typedef struct node
{
int data;
struct node *next;
}node;
int main()
{
int i, j, num, value;
node *p = NULL;
node *head = NULL;
printf("how many nodes\r\n");
scanf("%d",&num);
for(i = 0 ;i < num ; i++)
{
printf("enter node %d = ",i+1);
scanf("%d",&value);
p = (node *)malloc(sizeof(node));
p->data = value;
p->next = NULL;
// Form links
p->next = head;
head = p;
}
printf("linked list formed is \n");
for(p = head ; p != NULL ; p = p->next)
{
printf("%d ",p->data);
}
printf("\n");
// Freeing memory to avoid mem leaks
for(p = head ; head != NULL ; head = head->next)
{
p = head;
free(p);
}
return 0;
}
You can refer to my library for a more generic implementation of link_list

Answer 3

A change in code by Mohammed Meraj to create the list in the correct order.
#include <stdio.h>
#include <stdlib.h>
typedef struct node{
int data;
struct node *next;
}node;
int main()
{
int i, num, value;
node *p = NULL;
node *head = NULL;
node *q = NULL;
printf("how many nodes\r\n");
scanf("%d",&num);
for(i = 0 ;i < num ; i++)
{
printf("enter node %d = ",i+1);
scanf("%d",&value);
p = (node *)malloc(sizeof(node));
p->data = value;
p->next = NULL;
// Form links
if(!q) head = q = p;
else{ q->next = p; q = p; }
}
printf("linked list formed is \n");
for(p = head ; p != NULL ; p = p->next)
{
printf("%d ",p->data);
}
printf("\n");
// Freeing memory to avoid mem leaks
while(head != NULL)
{
p = head;
head = head->next;
free(p);
}
return 0;
}
if you want to do the code with only head and p, it can be done too
#include <stdio.h>
#include <stdlib.h>
typedef struct node{
int data;
struct node *next;
}node;
int main()
{
int i, num, value;
node *p = NULL;
node *head = NULL;
printf("how many nodes\r\n");
scanf("%d",&num);
for(i = 0 ;i < num ; i++)
{
printf("enter node %d = ",i+1);
scanf("%d",&value);
if(!head){
head = (node *)malloc(sizeof(node));
head->data = value;
head->next = NULL;
p = head;
}else{
p->next = (node *)malloc(sizeof(node));
p->next->data = value;
p->next->next = NULL;
p = p->next;
}
}
printf("linked list formed is \n");
for(p = head ; p != NULL ; p = p->next)
{
printf("%d ",p->data);
}
printf("\n");
// Freeing memory to avoid mem leaks
while(head != NULL)
{
p = head;
head = head->next;
free(p);
}
return 0;
}