Merrill Cornish wrote:

With an ordinary C++ pointer, if I have a BaseClass and a DerivedClass derived from BaseClass, then I can do

BaseClass* objectPtr; objectPtr = new DerivedClass();

Later, if BaseClass has a virtual destructor, I can call delete(objectPtr) and the DerivedClass destructor will be called.

Now, does this work with shared_ptr?

shared_ptr<BaseClass> objectPtr; objectPtr = shared_ptr<DerivedClass>(new DerivedClass());

and later, delete(objectPtr).

No, you shouldn't call delete directly. It's done for you automatically when the last shared_ptr that points to your object is destroyed.

Does any of the magic going on behind the scenes prevent shared_ptrs from being used like this?

shared_ptr would be close to useless if it didn't. In fact, shared pointer does *more* to remember the dynamic type than ordinary pointers. Forex, you can do this: { shared_ptr<void> voidPtr; voidPtr = shared_ptr

Merrill Cornish wrote:

With an ordinary C++ pointer, if I have a BaseClass and a DerivedClass derived from BaseClass, then I can do

BaseClass* objectPtr; objectPtr = new DerivedClass();

Later, if BaseClass has a virtual destructor, I can call delete(objectPtr) and the DerivedClass destructor will be called.

Now, does this work with shared_ptr?

shared_ptr<BaseClass> objectPtr; objectPtr = shared_ptr<DerivedClass>(new DerivedClass());

Yes, it does. It will work even without a virtual destructor.

and later, delete(objectPtr).

This doesn't work.

Does any of the magic going on behind the scenes prevent shared_ptrs from being used like this?

Of course not. :-)