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-... -Thorsten