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 19:58:04

vicente.botet wrote:
> ----- Original Message ----- From: "Joel de Guzman"
> <joel_at_[hidden]> To: <boost-users_at_[hidden]> Sent: Tuesday,
> January 06, 2009 11:26 PM Subject: Re: [Boost-users] [fusion] const and
> transform algorithm
> 
> 
>> 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.
> 
> Can for_each mutate the sequence elements? 

Yes.

if yes, would the following get a
> transformation I'm locking for?

I don't know. I'm not sure what you are looking for.

> vector<shared_pair<shared_future<int>,int>,shared_pair<shared_future<char>,int>
> > t; boost::fusion::for_each(t, intrusive_get()); vector<int, char>
> res=unzip<1>(boost::fusion::for_each(t, intrusive_get()));
> 
> where shared_pair would be a sequence something like a shared_ptr<pair<A,B>
> and unzip would be the reverse function of zip. BTW do you think that the
> counterpart function of zip has a place on Fusion?

I am not sure what it is you want to do. Could you explain in english
instead of non-functioning code?

>>> 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);
> 
> I was lock for in the documentation and I have not found this assignation.
> Where in the documentation can I found that I can do that?

See the sequence/views concepts and the docs for the containers.
For example (http://tinyurl.com/84vdvp):

Notation:
v Instance of vector
s A Forward Sequence

Expression Semantics:
  v =  s  Assigns to a vector, v, from a Forward Sequence, s.

> If for_each can change/mutate the sequence itself, why don't have a
> mutating_transform that takes a Sequence& and not a Sequence const&

Because it is not needed.

> Just a last question. How move semantics maps with purely functional
> Sequences? Do you plan to add move semantics for the elements of the tuple?
> The concept of MovableSequence has a sens for you?

Makes sense. Last BoostCon, we had a workshop about implementing
Fusion for 0x incuding move. We'll have it when it is available.

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