Re: [Boost-users] Is there a portable way to determine whether a type is functor ?

Subject: Re: [Boost-users] Is there a portable way to determine whether a type is functor ?
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2008-12-26 17:56:43

AMDG

ancode tran wrote:
> Hi all,  
>
> The following code only works under g++ 4.x:  
>
> #include <boost/type_traits.hpp> 
>
>  template <class T> 
> struct is_functor_helper 
> {  
>    static boost::type_traits::no_type test (...);  
>    template <class U>  
>    static boost::type_traits::yes_type test (U*, typeof(&U::operator()) dummy = NULL);  
>   BOOST_STATIC_CONSTANT (bool, value = sizeof(test((T*)NULL)) == sizeof(boost::type_traits::yes_type)); 
> };  
>
> template <class T> struct is_functor: public boost::integral_constant<bool, is_functor_helper<T>::value> { };   
>
> Can anyone give me a hint to make it work under other toolset, please ?  
> Thanks in advance.  
>   

That won't work with an overloaded or templated operator().

In general, there is no way to determine whether an operator() exists.
If you want to determine whether a type can be called with particular 
parameters see
http://www.boost.org/doc/html/proto/appendices.html#boost_proto.appendices.implementation.function_arity

In Christ,
Steven Watanabe