Re: [Boost-users] [enable_shared_from_this] My derivation ofenable_shared_from_this wants to be templated

Rob Currey:

template<class X> class Y: public enable_shared_from_this< Y<X> > { public:

shared_ptr< Y<X> > f() { return shared_from_this();

In a template, you need return this->shared_from_this(); to inform the compiler that shared_from_this will come from a base class.

} };