Re: Is it possible to insert multiple pointer to pointers in std::set<node **, iterator_pointer> ?

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Wed, 6 Jul 2011 20:20:53 CST
Am 06.07.2011 20:20, schrieb allswellthatendswell:

   We created a std:::set<node **, iterator_pointer> where node is
defined by this class: class node{
        Range value;
       //value stored in the node
       node *next;
       //pointer to next node
       node *prev;
       //pointer to previous node
       node(){ next = NULL; prev = NULL; }
       node(Range r){ value = r; next = NULL; prev = NULL; }

and iterator_pointer is defined by the struct: struct iterator_pointer
{ bool operator()(node** r1, node** r2) const{
    if (*r1 == 0 || *r2 == 0){
         return false;
                return ((*r1)->value.high()< (*r2)->value.high());

For some unknown reason when we only insert one pointer to pointer
(node**) into the std::set<node**, iterator_pointer>. When we try
inserting multiple unique pointer to pointers(node __), the size of
the std:set<node**, iterator_pointer> stays the same.
       We tried stepping into STL template code but we couldn't
determine why the size of std:Set does not continue increasing for
each successive unique node** insert. Are we doing something wrong or
is not possible to insert multiple unique keys correctly. Thank you.

Yes, you are doing something wrong: Your ordering criterion
iterator_pointer does not satisfy the requirements of a strict weak
ordering. In particular the first test

if (*r1 == 0 || *r2 == 0){
   return false;

is clearly defect, because it ensures that - given an object comp of
that type and an object n of type node** such that *n == 0 - both

comp(x, n) == false
comp(n, x) == false

evaluate to false, thus having the effect that x and n are equivalent
for *every* pointer value x. The obvious fix is to define a proper
ordering between null pointee and non-null pointee values, e.g.

bool operator()(node** r1, node** r2) const {
  if (*r1 == 0){
     return (*r2 != 0);
  } else if (*r2 == 0) {
     return false;
  return ((*r1)->value.high() < (*r2)->value.high());

HTH & Greetings from Bremen,

Daniel Kr?gler

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