Re: [boost] Giving names to formal parameters of lambda expressions: a named placeholder trick.

From: Giovanni Piero Deretta (gpderetta_at_[hidden])
Date: 2008-04-02 18:36:18

On Wed, Apr 2, 2008 at 11:59 PM, shunsuke <pstade.mb_at_[hidden]> wrote:
> Giovanni Piero Deretta wrote:
>  > Hum, if you want to have guaranteed static initialization, you need to
>  > statically know the function pointer value:
>  >
>  > template<typename Sig, Sig * p>
>  > static_function_ptr  {
>  >    typedef typename result_of_sig<Sig>::type result_type;
>  >
>  >    template<typename Args...>
>  >    result_type operator()(Args.... args) {
>  >        return p(args);
>  >    }
>  > };
>  >
>  > Substitute the pseudo-variadic syntax with proper preprocessor magic :).
>
>  Egg secretly has the feature.
>  See "egg/inlined.hpp".
>

Inline because it is practically the only way to get gcc to inline a
function pointer :) ?

Anyways, one reason less to need complex static initialization.

>
>  >>  I'm really afraid that those macros may keep users away from the library.
>  >>
>  >
>  > Hey, maybe it is just me, maybe others won't have any problems with
>  > them. I just find the macro based initialization ugly and unnecessary
>  > complex.
>
>  You expect something like this?:
>
>  static_result_of<T_curry2(F)>::type c = BOOST_EGG_CURRY2({}); // static

Can't you use boost::result_of here? (or return_of)
Also, do you have a plan for allowing complex statically initialized
expressions in header files
without ODR violations?

>  boost::result_of<T_curry2(F const &)>::type c = curry2(f); // dynamic

or egg::return_of if you want recursive evaluation.

>  stateless_result_of<T_curry2(F)>::type c = BOOST_EGG_STATELESS(); // for stateless one.

Why not:

   expression<T_curry2(F)> c = { /*just this please :) */ };  // note
the missing ::type

-- 
gpd