Subject: Re: [Boost-users] [boost][proto] Type inference using transform
From: joel falcou (joel.falcou_at_[hidden])
Date: 2010-05-06 05:13:05
Moreover, your result are wrong
result_of<T()> will always return void.
You need somethign like:
struct arg_tag
{};
namespace proto = boost::proto;
namespace mpl = boost::mpl;
using namespace std;
template<int N>
struct arg
: proto::or_<proto::nullary_expr<arg_tag, mpl::int_<N> > >
{};
template<typename T>
struct readonly
: proto::callable {
template<typename Sig>
struct result;
template<typename This,class X>
struct result<This(X)>
{
typedef const T& type;
};
};
template<typename T>
struct readwrite
: proto::callable {
template<typename Sig>
struct result;
template<typename This,class X>
struct result<This(X)>
{
typedef T& type;
};
};
template<int N, typename T>
struct argtype
: proto::or_<
proto::when<proto::assign<arg<N>, proto::_>, readwrite<T>(proto::_)>
, proto::otherwise<readonly<T>(proto::_)>
>
{};
proto::nullary_expr<arg_tag, mpl::int_<0> >::type const _1 = {{}};
proto::nullary_expr<arg_tag, mpl::int_<1> >::type const _2 = {{}};
namespace boost { namespace proto
{
template<class T> struct is_callable<readwrite<T> > : true_ {};
template<class T> struct is_callable<readonly<T> > : true_ {};
}}