Re: [boost] meaning of "unify"? (was Re: [Root Pointer] Seeking a Review Manager

Subject: Re: [boost] meaning of "unify"? (was Re: [Root Pointer] Seeking a Review Manager
From: Phil Bouchard (philippeb8_at_[hidden])
Date: 2016-04-04 18:22:57

On 04/04/2016 01:48 PM, Larry Evans wrote:
>
> Could you please explain how an intrusive list prevents cycles.

Wherever you want to insert an element inside an intrusive_list, it will 
automatically reorganize the pointers to the previous node and to the 
next node so that the former references are lost and the new ones are 
created.  You can have a cycle if former references are destroyed first.

But I might confuse you even more by trying to explain this 
extra-curriculum subject.

> I'm probably misinterpreting what the arrow means from the lower
> node_proxy to the upper node_proxy.  I was interpreting that to
> mean that node_proxy contained some node_proxy* member variable.
> However, looking at the code here:
>
> https://github.com/philippeb8/root_ptr/blob/master/include/boost/smart_ptr/root_ptr.hpp#L67
>
> and searching for node_proxy, I see no such member variable pointer.

So when a node_proxy is unified with another one it enlists the new 
node_proxy with the following tag:

mutable smart_ptr::detail::intrusive_list::node proxy_tag_;

You can see here that the way to iterate through all node_proxies 
unified together is done the following way:

using namespace smart_ptr::detail;

for (intrusive_list::iterator<node_proxy, &node_proxy::proxy_tag_> 
i(&proxy_tag_);;)
{
     ...

     if (++ i == intrusive_list::iterator<node_proxy, 
&node_proxy::proxy_tag_>(&proxy_tag_))
         break;
}

For the record, the list of allocated nodes associated to a specific 
node_proxy is stored in the following intrusive list:

mutable smart_ptr::detail::intrusive_list node_list_;

> Could you further explain the meaning of the arrow in that
> #root_ptr.rationale.union diagram?

Once again, when node_proxies are unified then it will use 
node_proxy::proxy_tag_ to link the node_proxies altogether.