linked list insert function - passing a list by pointer

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?

Sergey Kalinichenko · Accepted Answer · 2012-07-17 01:19:06Z

3

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.

answered Jul 17, 2012 at 1:19

Sergey Kalinichenko

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2 Comments

Taher Over a year ago

Could you please further explain the difference between the two cases?

Sergey Kalinichenko Over a year ago

@Smokie The first is an assignment of a pointer next itself to the pointer pointed to by l; the second one is an assignment of a pointer to next to the variable l itself.

Scott Hunter · Accepted Answer · 2012-07-17 01:18:50Z

1

Your function should only change *l if it is inserting into an empty list, as this is the only case where the first element of the list changes. This can be accomplished by using a local variable instead of *l inside of your function for those cases (initialized to *l).

answered Jul 17, 2012 at 1:18

Scott Hunter

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Comments

oalah · Accepted Answer · 2012-07-17 01:27:16Z

0

if you don't move the pointer l, then it remains at the head of the list. First assign l to temp, then move temp along the list but leave pointer l alone.

answered Jul 17, 2012 at 1:27

oalah

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