Larry Evans wrote:
On 03/29/2006 05:05 PM, Peter Dimov wrote:
make_shared_ptr<T>( a, b, c ) (or however we end up calling it) still has the advantage of being able to fold the two allocations into one.
I assume the two allocations are for the detached refcount and the referent. Wouldn't placing them in the same block of memory with a single call to new (and I assume this is what you mean by "fold the two allocations into one") mean that shared_ptr would have to change it's two deletes (1 for refcount and 1 for referent) into one?
Yes, the referent will be destroyed in place via explicit destructor call, not via delete.
IOW, make_shared_ptr would not make the current shared_ptr but another type of smart pointer. Or am I missing something?
The current shared_ptr can handle that.