remove the last_object from chain

also tried to keep the first around as was done in space
partially to not have to add the consrtants seperately
didn't work, as chain gets broken
also this has to be redone anyway

lib/parfait/factory.rb

@@ -13,7 +13,7 @@ module Parfait
   # factory when the next (not current) is nil.
   # This is btw just as easy a check, as the next needs to be gotten to swap the list.
   class Factory < Object
-    attr :type , :for_type , :next_object , :last_object , :reserve , :attribute_name
+    attr :type , :for_type , :next_object , :reserve , :attribute_name
 
     PAGE_SIZE = 1024
     RESERVE = 10
@@ -50,8 +50,8 @@ module Parfait
     # and rebuilt the reserve (get_next already instantiates the reserve)
     #
     def get_more
-      chain = get_chain
-      link = chain
+      first_object = get_chain
+      link = first_object
       count = RESERVE
       while(count > 0)
         link = get_next_for(link)
@@ -59,7 +59,8 @@ module Parfait
       end
       self.next_object = get_next_for(link)
       set_next_for( link , nil )
-      self.reserve = chain
+      self.reserve = first_object
+      self
     end
 
     # this initiates the syscall to get more memory.
@@ -111,6 +112,7 @@ module Parfait
       r_class = eval( "Parfait::#{type.object_class.name}" )
       obj = r_class.allocate
       obj.set_type(type)
+      #puts "Factory #{type.object_class.name} at 0x#{obj.object_id.to_s(16)}"
       obj
     end
   end