Linked Lists in C without malloc

Question

#include <stdio.h>

typedef struct node
{
      int i;
      struct node *next;
}node;

node getnode(int a)
{
      struct node n;
      n.i=a;
      n.next=NULL;
      return n;
}

main()
{
     int i;
     node newtemp,root,temp;

     scanf("%d",&i);
     root=getnode(i);
     temp=root;

     while(i--)
     {
         newtemp=getnode(i);

         temp.next=&newtemp;
         if(root.next==NULL)
         {
            root=temp;
         }
        temp=*(temp.next);
     }


     temp=root;

     while( temp.next != NULL )
     {
         printf(" %d ",temp.i);
         temp=*(temp.next);
     }
}

I am trying to create a linked-list without using malloc. The programming is printing only the root and no nodes following it. I couldn`t find the bug. Had there been any memory issue, the gcc compiler would have thrown a segmentation fault.(?) Please ignore the poor programming style..

Answer 1

When you initialise temp.next, what is the value of the pointer that you assign to it?

temp.next=&newtemp;

Why, it's the same one every time! (Draw a picture if you are unconvinced.)

Give it up. If you need an indeterminate amount of memory (which, with an indeterminate number of nodes, you do), then you need to allocate memory.