Explanation of code (linked list C)

Question

This is not my code. I took this code off this website:

http://www.macs.hw.ac.uk/~rjp/Coursewww/Cwww/linklist.html

I am using for reference material on how to build a linked list. I'm a little confused on what is going on. Can someone please explain to me what is going on. I'll mark what is confusing me with 1-5.

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

struct list_el {
   int val;
   struct list_el * next;
};

typedef struct list_el item;

void main() {
   item * curr, * head;
   int i;

   head = NULL;   //1

   for(i=1;i<=10;i++) {
      curr = (item *)malloc(sizeof(item));
      curr->val = i;
      curr->next  = head; //2
      head = curr; //3
   }

   curr = head; // 4

   while(curr) {  //5
      printf("%d\n", curr->val);
      curr = curr->next ;
   }

1. head = NULL → why is head being set to NULL? I know that you are supposed to (I do it out of habit) but I don't really know why.

2. curr->next = head → I never really understood this as well. Maybe I have my definition of "head" wrong but in a regular linked list, is it the starting node or the last node on the list? I've always assumed it was the starting node but in this line it looks like it's the last node.

3. head = curr → Why are we setting it equal to curr?

4. curr = head → and then setting curr = head after the loop is done.

5. while(curr) → Just to make sure, this is traversing through the list and it is equivalent to while(curr != NULL) right?

Answer 1

#1: head = NULL

Initializing the pointer. It's generally recommended to initialize the pointer to NULL either (1) at declaration or (2) immediately after declaration. If programmers mistakenly dereference uninitialized pointers, garbage values are returned. Often times, this is extremely hard to debug if your static analyzer and compiler don't display warning or error messages for uninitialized pointers.

For more information, please refer to Steve McConnell's Code Complete: A Practical Handbook of Software Construction or Wikipedia page on Defensive Programming.

#2: curr->next = head

Building the linked list. The curr node is being "linked" to previously created node in the sequence.

#3: head = curr

Updating the head pointer. The head pointer is being updated to point to the most recently malloced node.

Below illustrations visualize Steps #2 and #3:

#4: curr = head

Re-initializing the pointer. This step is similar to step #2:curr->next = head. By setting curr node to head, curr gets "ready" for linked-list traversal in the while loop. Analogically speaking, it's like initializing the iterating variable to 0 in the beginning of the loop (i.e. i = 0). To visualize this step, please refer to the below illustrations showing before/after this statement is executed:

#5: while(curr)

Traversing the list. Given that curr is pointing to the first node (from step #4), this while loop traverses the list until curr->next returns NULL. In a less abstract form, we can rewrite this statement as while(curr != NULL).