Unable to create and display linked list in C

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

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;
}