Subject: Re: [Boost-users] [lambda] extending lambda functions
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2009-03-24 11:48:17
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<norm_impl<R> > 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<std::complex<T> > {
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