The following code was been given from our school as a sample for circular linkedlist and told each student to build a own version of circular linkedlist.
My question is, is the following code really a circular linkedlist?
// Program of circular linked list
#include <stdio.h>
#include <malloc.h>
struct node
{
int info;
struct node *link;
}*last;
main()
{
int choice,n,m,po,i;
last=NULL;
while(1)
{
printf("1.Create List\n");
printf("2.Add at begining\n");
printf("3.Add after \n");
printf("4.Delete\n");
printf("5.Display\n");
printf("6.Quit\n");
printf("Enter your choice : ");
scanf("%d",&choice);
switch(choice)
{
case 1:
printf("How many nodes you want : ");
scanf("%d",&n);
for(i=0; i < n;i++)
{
printf("Enter the element : ");
scanf("%d",&m);
create_list(m);
}
break;
case 2:
printf("Enter the element : ");
scanf("%d",&m);
addatbeg(m);
break;
case 3:
printf("Enter the element : ");
scanf("%d",&m);
printf("Enter the position after which this element is inserted : ");
scanf("%d",&po);
addafter(m,po);
break;
case 4:
if(last == NULL)
{
printf("List underflow\n");
continue;
}
printf("Enter the number for deletion : ");
scanf("%d",&m);
del(m);
break;
case 5:
display();
break;
case 6:
exit(0);
default:
printf("Wrong choice\n");
}/*End of switch*/
}/*End of while*/
}/*End of main()*/
create_list(int num)
{
struct node *q,*tmp;
tmp= malloc(sizeof(struct node));
tmp->info = num;
if(last == NULL)
{
last = tmp;
tmp->link = last;
}
else
{
tmp->link = last->link; /*added at the end of list*/
last->link = tmp;
last = tmp;
}
}/*End of create_list()*/
addatbeg(int num)
{
struct node *tmp;
tmp = malloc(sizeof(struct node));
tmp->info = num;
tmp->link = last->link;
last->link = tmp;
}/*End of addatbeg()*/
addafter(int num,int pos)
{
struct node *tmp,*q;
int i;
q = last->link;
for(i=0; i < pos-1; i++)
{
q = q->link;
if(q == last->link)
{
printf("There are less than %d elements\n",pos);
return;
}
}/*End of for*/
tmp = malloc(sizeof(struct node) );
tmp->link = q->link;
tmp->info = num;
q->link = tmp;
if(q==last) /*Element inserted at the end*/
last=tmp;
}/*End of addafter()*/
del(int num)
{
struct node *tmp,*q;
if( last->link == last && last->info == num) /*Only one element*/
{
tmp = last;
last = NULL;
free(tmp);
return;
}
q = last->link;
if(q->info == num)
{
tmp = q;
last->link = q->link;
free(tmp);
return;
}
while(q->link != last)
{
if(q->link->info == num) /*Element deleted in between*/
{
tmp = q->link;
q->link = tmp->link;
free(tmp);
printf("%d deleted\n",num);
return;
}
q = q->link;
}/*End of while*/
if(q->link->info == num) /*Last element deleted q->link=last*/
{
tmp = q->link;
q->link = last->link;
free(tmp);
last = q;
return;
}
printf("Element %d not found\n",num);
}/*End of del()*/
display()
{
struct node *q;
if(last == NULL)
{
printf("List is empty\n");
return;
}
q = last->link;
printf("List is :\n");
while(q != last)
{
printf("%d ", q->info);
q = q->link;
}
printf("%d\n",last->info);
}/*End of display()*/
The reason why I am not agreeing is because NULL is used to check the last node in the list.
Yes, the code implements a circular linked list
The variable last is only NULL at the beginning, after adding the first element lasts link will point to itself. See function createlist.
Non circular lists always have the last elements next pointer set to NULL to indicate the end of the list.
Edit:
Sorry I can’t do ascii art with my ipad.
To understand how these pointers work I would recommend to draw a simple diagramm
based on the instructions in the code. Hope this helps.