Question

To check a sample list imlpementation I tried the following code. But whenever i am trying to display the result it is getting inside a loop. I cant find where is it getting wrong.

#include<stdio.h>
#include<stdlib.h>

typedef struct linkedlist
{
    int data;
    struct linkedlist *next;
}node;

int main()
{
    int ch,num;

    node *head=NULL;
    head=(node *)malloc(sizeof(node));
    node *new=NULL;
    new=(node *)malloc(sizeof(node));
    node *temp=NULL;
    temp=(node *)malloc(sizeof(node));
    printf("\n1.Insert to list");
    printf("\n3.Display the list");
    printf("\n Enter Choice->");
    scanf("%d",&ch);
    switch(ch)
    {
     case 1:printf("\n Enter data->");
            scanf("%d",&num);
            new->data=num;
            new->next=NULL;
            head->next=new;
            break;

     case 3: temp=head;
            while(temp!=NULL)
            {
                printf("\n %d",temp->data);
                temp=temp->next;
            }


                break;
     default:printf("Wrong Choice");
              break;

    }
    return 0;
 }

Answer 1

There are two mistakes here.

1. You are only allocating “new” once. This means you are reusing the same node every time the user enters “1” and actually linking it to itself.
2. You shouldn’t be allocating a node for “temp” at all, since you are not using it. You are actually loosing the pointer to that allocated node with this line: case 3: temp=head; This is called a “memory leak.”

I suggest that you study pointers a bit more; they seem to be confusing you.