Re: [Boost-users] [lambda] extending lambda functions

24 Mar 2009


      AMDG

alfC wrote:
...
...
result is a template, so you just need to specialize it or use
some kind of metaprogramming (for instance Boost.Typeof).
Do you mean that I should do something like (not tested)::
  template<class R>
  struct norm_impl{
    template <typename Arg>
    struct result{
      typedef R type;
    };
    template <typename Arg>
    typename result<Arg>::type operator()(Arg ar1) const{
      return std::norm(ar1);
    }
  };
  template<class R> struct norm{
  static(???) phoenix::function func;
};
and use norm<double>::func (instead of norm_);
How Boost.TypeOf fits in this or other approach?
No.  I meant something like this:

struct norm_impl {
    template<typename Arg>
    struct result {
        typedef Arg type;
    };
    template<typename T>
    struct result {
        typedef T type;
    };
    //...
};

Using Boost.Typeof result could be

template<class Arg>
struct result {
    BOOST_TYPEOF_NESTED_TYPEDEF_TPL(nested, (std::norm(Arg())));
    typedef typename nested::type type;
};
...
...
...
2) I may be asking for too much sugar but is there a way to name the
function "std::norm" and not have conflicts with the standard
"std::norm<T>(std::complex<T> const& c)", the best I could do was to
call it std::norm_;
You're not allowed to put it in namespace std at all.
Obviously there is a mapping between standard functions and lazy
functions (in Phoenix for example), Is there a way to express this
mapping in their names?
Just put them in your own namespace.
The ones defined by phoenix are in namespace
boost::phoenix for instance.

In Christ,
Steven Watanabe