Re: [Boost-users] [ptr_container] ownership

From: Thorsten Ottosen (thorsten.ottosen_at_[hidden])
Date: 2006-12-09 11:17:44

David Grüner wrote:
> Hello,
> i need to give up the ownership of a ptr_container (vector here)
> over its pointers (without assignment to a new ptr_container).
> Two questions here:
> 
> 1. Do i really have to write aPtrContainer.release().release()?
> It looks weird. For me, the function name release() of the container
> seems confusing since an unassigned aPtrContainer.release() just
> does the "opposite" of auto_ptr<T>::release(), namely deleting the
> pointers. ;o)

Right, there is a difference between a single smart pointer and a 
container here.

Even if release() in a pointer container returned ptr_container*,
it would still not be consistent with auto_ptr<T>, since that class 
returns T*, but ptr_container<T>, would return ptr_container<T>*.

But one could say, why not simply let release() return
container<T*>? That function cannot be code more efficiently as a member 
function (due to the over encapsulated interface of eg. std::vector).
You have many options here, say

template< class PtrSequence >
std::vector< typename PtrSequence::value_type >
release( PtrSequence& seq )
{
   std::vector< typename PtrSequence::value_type > result;
   while( !seq.empty() )
      result.push_back( seq.pop_back().release() );
   std::reverse( result.begin(), result.end() );
}

> The behavior is more like transfer_ownership()
> or something similar ($0.02).
> 
> 2. Beside managing memory, i plan to use ptr_container primarily to
> propagate const. So, in some cases i need a ptr_container which just
> doesn't own any given memory. What's the best way to do that?

use a non-owning clonse_allocator:

http://www.boost.org/libs/ptr_container/doc/reference.html#class-view-clone-allocator

-Thorsten