Re: [boost] [type_traits?]enable_if an overload for a function exists

From: Steven Watanabe (steven_at_[hidden])
Date: 2007-04-19 22:26:04

AMDG

Marco <mrcekets <at> gmail.com> writes:

> On Thu, 19 Apr 2007 23:10:31 +0200, Michael Marcin  
> <mmarcin <at> method-solutions.com> wrote:
> 
> > I want to enable a function templated on a type T if a function it  
> > relies on overloaded on that type.
> >

How about?

template<class T>
yes is_overloaded(T);
template<class T>
no is_overloaded(...);

#define BOOST_IS_OVERLOADED(t, name)                 \
    (sizeof(is_overloaded<t>(name)) == sizeof(yes))

No that doesn't work either.  If there are multiple
functions the compiler won't know which one to pass
to is_overloaded(...)

Um.  Since you are looking for the exact
signature (untested):

namespace detail {

namespace is_overloaded_adl_barrier {

typedef chr no;
struct yes { no dummy[2]; };

struct no_overload {};

yes is_overloaded_impl(...);
no is_overloaded_impl(const no_overload&);
struct any_conversion {
    template<class T>
    any_conversion(const T&);
};

no_overload atan2(any_conversion, any_conversion);

template<class T>
T make();

template<class T>
struct has_atan2 :
  mpl::bool_<
    sizeof(is_overloaded_impl(atan2(make<T>(), make<T>()))) ==
    sizeof(yes)
  >
{};

}

using is_overloaded_adl_barrier::has_atan2;

}

In Christ,
Steven Watanabe