C program to free the nodes of a linked list

Before looking at the answer, try writing a simple C program (with a for loop) to do this. Quite a few people get this wrong.


This is the wrong way to do it

struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = listptr->next)
{
  free(listptr);
}


If you are thinking why the above piece of code is wrong, note that once you free the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is already freed, using it to get listptr->next is illegal and can cause unpredictable results!



This is the right way to do it


struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = nextptr)
{
  nextptr = listptr->next;
  free(listptr);
}
head = NULL;


After doing this, make sure you also set the head pointer to NULL!