Re: [boost] [type_erasure] any and shared_ptr

16 Oct 2012


      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