list::merge and GLIBCXX_DEBUG

[David Greene]

I hit something in GLIBCXX_DEBUG mode the other day and I
want to ask about it.

Do list iterators remain valid after splice and merge operations?

The code is below.  Is this a valid program?  If not, what assumptions
am I violating?

                                   -Dave

#include <list>
#include <iostream>

int main(void)
{
   std::list<int> list1;
   std::list<int> list2;

   for(int i = 0; i < 10; ++i) {
     list1.push_back(i);
     list2.push_back(10-i);
   }

   list1.sort();
   list2.sort();

   std::list<int>::iterator node_of_interest = list2.begin();

   // Ok, preserves iterator and updates iterator owner in
   // GLIBCXX_DEBUG mode
   list1.splice(list1.begin(), list2, node_of_interest);
   // Ok, preserves iterator and updates iterator owner in
   // GLIBCXX_DEBUG mode
   list2.splice(list2.begin(), list1, node_of_interest);

   // Oops, does not appear to update iterator owner in GLIBCXX_DEBUG
   // mode
   list1.merge(list2, std::less<int>());

   // Triggers GLIBCXX_DEBUG error: "attempt to splice an iterator from
   // a different container."  Is node_of_interest still a valid
   // iterator according to the standard?
   list2.splice(list2.begin(), list1, node_of_interest);

   std::cout << "list1:\n";
   for(std::list<int>::iterator i = list1.begin();
       i != list1.end();
       ++i) {
     std::cout << *i << "\n";
   }

   std::cout << "list2:\n";
   for(std::list<int>::iterator i = list2.begin();
       i != list2.end();
       ++i) {
     std::cout << *i << "\n";
   }

   return 0;
}