Re: [boost] Detecting smart pointers: is_deferencable, a solution?

From: John Maddock (john_at_[hidden])
Date: 2007-01-22 05:16:04

Daniel Wallin wrote:
> John Maddock wrote:
>> Dmitry Ivankov wrote:
>>> Doesn't work for void operator*(B&), but can we call such class
>>> dereferencable? :)
>>
>> No, I think there will always be some corner cases like that where
>> it blows up, but does anyone really write such an operator in
>> practice?
>
> It's quite possible to handle the case where *x is void. See for
> example boost/detail/is_incrementable.hpp. A couple of years ago, I
> tried to generalize this into something reusable and more powerful.
> The result of that is available in the sandbox;
> boost/result_check.hpp. The intended usage was something like:
>
>  BOOST_RESULT_CHECK(1, operator*, dereference, *_1)
>                     ^      ^           ^        ^
>           arity ----'      |           |        |
>                  function -'           |        |
>                     metafunction name -'        |
>                                      operation -'
>
>  template <class T>
>  struct is_dereferencable
>    : check_dereference<T>
>  {};
>
>  template <class T>
>  struct is_dereference_void
>    : check_dereference<T, boost::is_void<mpl::_> >
>  {};
>
>  template <class T, class U>
>  struct is_dereference_convertible_to
>    : check_dereference<T, boost::is_convertible<mpl::_, U> >
>  {};

Very interesting, I've had a play with that, and it does indeed seem to be a 
superior solution.

Thanks!

John.