Re: [Boost-users] using lambda's pointer tomemberfunctionandboost::function

You are right, it doesn't work; lambda's operator->* seems a pretty odd beast. Interesting that nobody has discovered its quirks so far. FWIW, the syntax above is how it should work in order to be usable. Lambda currently implements

That's how I thought it would work. Is it difficult to make lambda work this way?

You can use the above syntax with boost::bind and overload get_pointer for aux, but this won't work for lambda::bind or std::tr1::bind. Can't you just add operator* to aux? A raw pointer can escape via ptr, so adding a way to dereference the pointer doesn't seem to make it any more dangerous.

No, I can't add operator* to aux because its ptr member function actually returns a shared_ptr that it gets from another source. If this source doesn't own the shared_ptr, ptr will return a shared_ptr with use_count 1. If we define this member as T &operator*() { return *ptr(); } it would return a reference to an object that is already deleted by ~shared_ptr. By using only ->* or -> returning a shared_ptr, it will last until the end of the member function call.

It works for your example and returns 5. When I add an aux_ptr class with a

My example doesn't compile with g++ 4.1.1, which compiler are you using?

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function f = &aux::a;

works through it, too.

What if aux::a has a parameter? I'd like to do: function f = (&_1->*&aux::a)(x,y,z); Thanks, Rodolfo Lima.