Can I use the type of "this" as a template argument (inside a macro)?

Question

I currently have this:

#define THIS(T) (boost::static_pointer_cast<T>(shared_from_this()))

The macro is used in a method like this:

void Derived::run() {
    do_something_with_a_shared_ptr(THIS(Derived));
}

That's all well and good, but I'd like to eliminate (Derived) and have:

void Derived::run() {
    do_something_with_a_shared_ptr(THIS);
}

Is this possible?

Alternatively, is there a better way to have convenient access to a shared_ptr to this, in a class derived (indirectly) from boost::enable_shared_from_this? This question seems to indicate the answer to this is no.

The class hierarchy looks like this:

class Base: public boost::enable_shared_from_this<Base> {
    ...
}

class Derived: public Base {
    ...
    void run();
    ...
}

void do_something_with_a_shared_ptr(boost::shared_ptr<Derived>);

Answer 1

Not exactly an answer to your question, but have you considered using a member function instead of a macro? I usually do something like:

boost::shared_ptr< Derived > shared_this()
{
     return boost::static_pointer_cast<Derived>(shared_from_this());
}
boost::shared_ptr< Derived const > shared_this() const
{
     return boost::static_pointer_cast<Derived const>(shared_from_this());
}