Do you have to pass delete the same pointer that was returned by new, or can you pass it a pointer to one of the classes base types? For example:
class Base
{
public:
virtual ~Base();
...
};
class IFoo
{
public:
virtual ~IFoo() {}
virtual void DoSomething() = 0;
};
class Bar : public Base, public IFoo
{
public:
virtual ~Bar();
void DoSomething();
...
};
Bar * pBar = new Bar;
IFoo * pFoo = pBar;
delete pFoo;
Of course this is greatly simplified. What I really want to do is create a container full of boost::shared_ptr and pass it to some code that will remove it from the container when it is finished. This code will know nothing of the implementation of Bar or Base, and will rely on the implied delete operator in the shared_ptr destructor to do the right thing.
Can this possibly work? My intuition says no, since the pointers will not have the same address. On the other hand, a dynamic_cast<Bar*> should work, so somewhere the compiler is storing enough information to figure it out.
Thanks for the help, everybody who answered and commented. I already knew the importance of virtual destructors, as shown in my example; after seeing the answer I gave it a little thought, and realized the whole reason for a virtual destructor is this exact scenario. Thus it had to work. I was thrown by the absence of a visible means of converting the pointer back to the original. A little more thinking led me to believe there was an invisible means, and I theorized that the destructor was returning the true pointer for delete to release. Investigating the compiled code from Microsoft VC++ confirmed my suspicion when I saw this line in ~Base:
mov eax, DWORD PTR _this$[ebp]
Tracing the assembler revealed that this was the pointer being passed to the delete function. Mystery solved.
I've fixed the example to add the virtual destructor to IFoo, it was a simple oversight. Thanks again to everyone who pointed it out.