Re: [Boost-users] [fusion] const and transform algorithm

Subject: Re: [Boost-users] [fusion] const and transform algorithm
From: Joel de Guzman (joel_at_[hidden])
Date: 2009-01-06 17:26:52

vicente.botet wrote:
> Hi,
> 
> What is the reason fusion::transform taskes a const Sequence?
> template<
>     typename Sequence,
>     typename F
>     >
> typename result_of::transform<Sequence const, F>::type transform(
>     Sequence const& seq, F f);
> 
> Does this means that the transform do not change the sequence itself o also its elements?

Yes. Fusion transform returns a lazy (transform) view. This view
is non-mutating. It is not like STL's transform which mutates
the target container in place. Fusion's is purely functional and
lazily evaluated.

> Why the following do not compiles
>     struct get {        
>         template<typename Sig> struct result;
>         template<typename T> struct result<get(T)> {            
>             typedef typename boost::remove_reference<T>::type ACT;            
>             typedef typename ACT::result_type type;        
>         };                    
>         template<typename ACT>        
>         typename ACT::result_type operator()(ACT& act) const 
>         {            return act.get();        }    
>     };
>     boost::tuple<shared_future<int>,shared_future<char> > t;
>     boost::fusion::result_of::transform<
>         boost::tuple<shared_future<int>,shared_future<char> > , get>::type res=
>             boost::fusion::transform(t, get());
> 
> transform call to get()(const shared_future<int>&act). How can I make a transformation changing also the sequence itself?
> 
> The result of transform is a transform_view, so the function get is not really called. How can I assign this to a 
>     boost::tuple<int, char> = ??(boost::fusion::transform(t, get()));
> boost::as_vector? Is there a as_tuple function? Should this force the evaluation of the get function? 

As I said, Fusion transform is purely functional. The result of a transform
is a view. The function will only be called when you iterate over the
view. The transformation does not change/mutate the sequence itself.
If that's what you want, you can assign the view back to the original
container. For example:

     vector<int, char> v;
     v = transform(v, f);

Regards,
-- 
Joel de Guzman
http://www.boostpro.com
http://spirit.sf.net