foo = boost::shared_ptr<Foo>(new Bar); foo = boost::shared_ptr<Bar>(new Bar); Scott McMurray wrote:

The important difference is in the default deleter. shared_ptr<Foo>'s default deleter will call delete on a Foo* while shared_ptr<Bar>'s default deleter will call it on a Bar*. Since your base class's destructor is virtual, it doesn't make any difference in the example. If your destructor were non-virtual, then you'd have to use shared_ptr<Bar>(new Bar) to get the correct destructors called. (But without any virtual functions, the dynamic casting wouldn't compile.)

Are you sure about that? A check of the documentation lists the signature for that constructor as template<class Y> explicit shared_ptr(Y * p); The explanation is given as follows: [This constructor has been changed to a template in order to remember the actual pointer type passed. The destructor will call delete with the same pointer, complete with its original type, even when T does not have a virtual destructor, or is void. The reset function has a similar signature.