Since you're not calling shared_from_this() in A, I suggest you remove
enable_shared_from_this<A> altogether.
But in my case there is no option like this (Class A is a base library
class). So I have to do this.
void f(void)
{
cout << "this is working " <<
boost::enable_shared_from_this<B>::shared_from_this() << endl;
}
};
You create an empty enable_shared_from_this<B>, then call shared_from_this()
on that one which is of course empty.
I did the above to avoid the ambiguous call error
So how can I specify that i am calling the shared_from_this() in class
B....?
On Mon, Aug 10, 2009 at 4:33 AM, Christian Holmquist
void f(void)
{ cout << "this is working " << boost::enable_shared_from_this<B>::shared_from_this() << endl; } };
You create an empty enable_shared_from_this<B>, then call shared_from_this() on that one which is of course empty. Intended usage is
class B : public A, public boost::enable_shared_from_this<B> { public: B() {} virtual ~B() {}
void f(void) { cout << "this is working " << shared_from_this() << endl; // ^^^^^^^^^^^^^^^^^^^^^^ } };
This is not the only error though, Since you're deriving from enable_shared_from_this two times, the above will give an ambiguous call error (did you mean enable_shared_from_this<A> or enable_shared_from_this<B>).
Since you're not calling shared_from_this() in A, I suggest you remove enable_shared_from_this<A> altogether.
/ christian
_______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users
_______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users
-- Regards, R. P. Janaka