Question

Im trying to create linked-list insert function that takes a list (or more correctly a pointer to it) then inserts the value to the end of the list.

void ll_insert(struct ll **l, int n){
  struct ll *temp=NULL;
  while ( (*l) != NULL){
    temp= (*l);
    (*l) = (*l)->next;
  }
  (*l)= (struct ll*)malloc(sizeof(struct ll));
  (*l)->n=n;
  (*l)->next=NULL;
  if (temp) temp->next= (*l);
}


int main(void){
  struct ll *l=NULL;                                                                                                                                                         
  ll_insert(&l, 1);
  printf("%d ", l->n);
  ll_insert(&l, 1);
  ll_insert(&l, 2);
  ll_insert(&l, 3);
  printf("%d ", l->n); 

}

The output after running the code above is 1 3. This is no surprise, since

(*l) = (*l)->next;

updates the list to point to to the end node, and every time I run insert(…) the list’s head is updated to point to the end (if im not wrong). What’s the way around this?

Answer 1

You are not using the pointer to pointer correctly: this line in the while loop

(*l) = (*l)->next;

should be

l = &((*l)->next);

If you use it that way, you wouldn’t need your temp variable at all.

Since this is C, not C++, it is OK to not cast malloc.