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From: Joel de Guzman (joel_at_[hidden])
Date: 2007-10-07 19:15:17

Marco wrote:
> On Sun, 07 Oct 2007 10:41:52 +0200, Marco Costalba <mcostalba_at_[hidden]>
> wrote:
>> I would think to fix completely the problem reported by Joel we could
>> write instead:
>> boost::is_convertible<F, boost::function<T> >
>> But for unknown reasons this fails! All types of function pointers match!
>> Could someone please be so kind to enlight me?
>> Thanks
>> Marco
> Because boost::function has a constructor whose behavior is like this:
> template< typename Functor >
> function (Functor const& f)
> it skips only integral type through enable_if.
> So, except integral, everything is convertible to boost::function even
> apples.
> In order to implement signature deduction I exploits a support utility
> used in the implementation of boost::function and a lot of metaprogramming.
> Give a glance to detail::function::get_function_tag in function_base.hpp,
> in order to learn how to discern between different kind of functors.

Tobias (Schwinger) has this excellent library for dealing with function
detection: Boost.function_types. It's in SVN now...

However, I do not think this will help at all. Ok, so, say we have
a polymorphic function (object) called pf. Then we have an overload
set with 5 signatures, ov. pf can deal with 3 of them (say the first
three). Now, how do you automatically "assign" pf to the first
three overloads? You can't. There's no way to detect the signature
of a template function (the operator()):

Here's what I mean. Using the "add" syntax:

     ov.add(pf); // which one? duh!


Joel de Guzman

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