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 wrote:

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

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-- Regards, R. P. Janaka

Don't let B inherit from enable_shared_from_this. You've already done that in A. So your question then becomes "how do I retrieve a shared_ptr<B> from this?". Here's how:

 boost::shared_dynamic_cast(shared_from_this());

Why not: boost::static_pointer_cast<B>(shared_from_this());

Yeah.. exactly this is what I wanted to know.. thank you. 2009/8/10 Björn Karlsson

Hello R.P,

Inheriting from enable_shared_from_this more than once is not a good idea.

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From: boost-users-bounces@lists.boost.org [mailto:boost-users- bounces@lists.boost.org] On Behalf Of R. P. Janaka Sent: Monday, August 10, 2009 6:54 AM To: boost-users@lists.boost.org Subject: Re: [Boost-users] How shared_from_this() work when it inherited from more than one time

But in my case there is no option like this (Class A is a base library class). So I have to do this.

Don't let B inherit from enable_shared_from_this. You've already done that in A. So your question then becomes "how do I retrieve a shared_ptr<B> from this?". Here's how:

boost::shared_dynamic_cast(shared_from_this());

In the example you've provided, it would also be safe to use boost::shared_static_cast. To conclude: You should only inherit from enable_shared_from_this in A, and B should only inherit from A.

...

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....?

You did. But the problem is that you have two weak_ptrs -- one weak_ptr<A> and one weak_ptr<B> -- and only one of them will be initialized correctly when you create the smart pointer. (If you're interested in debugging your case to really understand what happens, have a look at the function sp_enable_shared_from_this() in shared_ptr.hpp.)

Does that help?

Cheers, Bjorn Karlsson

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-- Regards, R. P. Janaka

Hello Frank,

The shared_*_cast names are deprecated according to shared_ptr.hpp. They are spelled *_pointer_cast now.

You're right, thanks for the reminder. The rationale for this name change can be found in the smart pointer proposal, N1540: "The cast operations were originally named shared_static_cast and shared_dynamic_cast in the Boost implementation, but it soon became clear that other smart pointer types can (and should) provide the same functionality, preferably in a generic manner, and the functions were renamed to *_pointer_cast to indicate wider applicability." Cheers, Bjorn Karlsson www.skeletonsoftware.net