I'd like to request the following functions added to shared_ptr's interface:
template <class T> void * shared_ptr<T>::void_cast() const;
I'd rather spell that 'pointer()' or something along those lines for consistency with 'type()'.
I like pointer() better, yes.
template <class T> std::type_info const & shared_ptr<T>::type() const; <snip> But consider the following slightly more interesting example:
struct Base { virtual ~Base() {} };
struct Derived: Base { int i; };
Derived * pd = new Derived; Base * pb = pd; shared_ptr<void> pv( pb );
Now there are two reasonable interpretations; either
pv->pointer() == pb && pv->type() == typeid(Base);
or
pv->pointer() == pd && pv->type() == typeid(Derived);
Which one should shared_ptr pick?
I had in mind the second interpretation from your example, that is, ::type() would return the "object type" as defined in [1.8.1]. There is also the case when a shared_ptr is initialized with a pointer to incomplete (or void) type and a custom deleter, when the "object type" is unavailable for shared_ptr to pick up automatically. Therefore, if ::type() and ::pointer() are supported, it also makes sense to provide constructor(s) that take the std::type_info const & to be returned by ::type(). This way someone could presumably achieve the first interpretation from your example: Derived * pd = new Derived; Base * pb = pd; shared_ptr<void> pv( pb, typeid(Base) ); Emil Dotchevski