Question

I have found this piece of code in a book:

void DeleteList(element *head)
{
  element *next, *deleteMe;
  deleteMe = head;
  while (deleteMe) {
    next = deleteMe->next;
    free(deleteMe);
    deleteMe = next;
  }
}

Assuming that the argument that we are passing to the function is the pointer to the head of the list, why are we not passing a reference to that pointer?

If we don’t do that, aren’t we just going to delete a local copy of that pointer? That is why we pass reference pointers right? So the callee get the changes that have been done inside the function?

Answer 1

If we don’t do that, aren’t we just going to delete a local copy of that pointer?

Well, there is no such thing as a reference in C, but I assume you mean a pointer-to-pointer, and the answer would be no.

You are freeing the memory that the pointer points to on the heap, not the pointer itself. You are not mutating the pointer, so a copy is fine. Regardless of whether it is a copy or not, they point to the same place.

You are confusing this with assigning a completely new value to a variable passed to a function, in which case you would need to take a pointer-to-pointer. for example:

// only changes the local copy
void init(some_type *foo) {
    foo = malloc(sizeof(some_type));
}

// initializes the value at *foo for the caller to see
void init(some_type **foo) {
    *foo = malloc(sizeof(some_type));
}