Re: [boost] [functional] Interested in can_be_called<F, Sig> metafunction?

Subject: Re: [boost] [functional] Interested in can_be_called<F, Sig> metafunction?
From: Mathias Gaunard (mathias.gaunard_at_[hidden])
Date: 2012-11-06 18:10:46

On 04/11/2012 20:54, Matt Calabrese wrote:
> On Sun, Nov 4, 2012 at 2:23 PM, Mathias Gaunard <
> mathias.gaunard_at_[hidden]> wrote:
>
>> Personally, I've found it more useful to just not define
>> result_of<F(args)>::type if F(args) cannot be called.
>>
>
> The only problem with that is that you have to rely on the creator of a
> function object to abide by this convention. "can_be_called" (callable
> might be a simpler name and is what was used with C++0x concepts) would
> work regardless of whether or not result_of is defined.

result_of does not have to be implemented in terms of 
result<Sig>::type/result_type.

template<class T, class R = void>
struct enable_if_type
{
   typedef R type;
};

template<class Sig, class Enable = void>
struct result_of {};

template<class F, class... Args>
struct result_of<F(Args...)
                , typename enable_if_type<
                    decltype( std::declval<F>()
                              (std::declval<Args>()...)
                            )
                  >::type
               >
{
   typedef decltype(std::declval<F>()(std::declval<Args>()...)) type;
};

Of course, if you don't have decltype, you can't do that and must use 
the result_type or result templates.

In C++03, you'll be able to deduce whether the expression in valid, but 
not what the result type is.