Re: [boost] [type_erasure] any and shared_ptr

Subject: Re: [boost] [type_erasure] any and shared_ptr
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2012-10-16 12:08:10

AMDG

On 10/16/2012 07:52 AM, Christophe Henry wrote:
> Hi Steven,
> 
> I need to use the associated types but don't get them to do what I
> need. Consider
> 
> struct Pointee : public boost::enable_shared_from_this<Pointee>
> {
>   void foo()
>   {
>     std::cout << "Pointee::foo called" << std::endl;
>     //shared_from_this();
>     std::cout << "Pointee::foo end" << std::endl;
>   }
> };
> 
> Pointee needs to be enable_shared_from_this for some reason, which
> forces me to use it in a shared_ptr. I reused your pointee / pointer
> structures:
> 
> <snip>
> 
> I (naively) tried to define a pointee concept:
> 
> typedef mpl::vector<
>       pointer<>,
>       same_type<pointer<>::element_type,_a >,
>       relaxed_match,
>       has_foo<void()>
>     > PointeeConcept;
> 
>   // will not compile, shared_ptr<Pointee> has no foo...
>   boost::shared_ptr<Pointee> p (new Pointee) ;
>   any<PointeeConcept> ap (p);
>   ap.foo();
> 

The placeholder _self maps to shared_ptr<Pointee>
and the placeholder _a maps to Pointee.
has_foo<void()> is actually has_foo<void(), _self>
because of the default argument.  This says
that shared_ptr<Pointee> has a member called foo.
What you really want is has_foo<void(), _a> to
say that foo is a member of Pointee.

In Christ,
Steven Watanabe