Re: [Boost-users] Problems using mpl::if_ and function_types::result_type

Subject: Re: [Boost-users] Problems using mpl::if_ and function_types::result_type
From: Jeffrey Lee Hellrung, Jr. (jeffrey.hellrung_at_[hidden])
Date: 2011-11-24 14:38:35

On Thu, Nov 24, 2011 at 5:25 AM, Allan Nielsen <a_at_[hidden]> wrote:

> Hi
>
> I'm new to the boost::mpl library, and have some "beginners-problems"
>
> template < typename F >
> struct A {
>    typedef boost::function_types::parameter_types<F> P;
>    typedef typename boost::function_types::result_type<F>::type R;
>
>    typedef typename boost::mpl::if_< boost::is_same< R, void >,
>                                      boost::function< void ( void ) > ,
>                                      boost::function< void ( R ) > >::type
> TTT;
>    A() {
>    }
> };
>
> int main(int argc, const char *argv[])
> {
>    A<int(int, float)>  ok; // working
>    A<void(int, float)> compile_error; // non-working
>
>    return 0;
> }
>
>
>
> xxx.cxx: In instantiation of ‘A<void(int, float)>’:
> xxx.cxx:108:25:   instantiated from here
> xxx.cxx:100:77: error: invalid parameter type
> ‘boost::mpl::aux::wrapped_type<boost::mpl::aux::type_wrapper<void>
> >::type’
> xxx.cxx:100:77: error: in declaration ‘A<F>::TTT’
>
>
> What is the problem here, and how can I solve it.
>
> To my understanding only the selected part of mpl::if_ should be
> evaluated by the compiler....
>

I believe Max Vasin's conclusion is correct.  To be explicit, you're still
forming the function type "void ( R )", regardless of whether R is void or
not, and (I'm guessing here) the compiler's not cool with that if R is
void.  The easier replacement for your mpl::if_ metaexpression might just
be to specialize a helper struct:

template< class R > struct helper { typedef void type ( R ); };
template<> struct helper< void > { typedef void type ( ); };

HTH,

- Jeff