[fusion] const and transform algorithm

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vicente.botet

6 Jan 2009 6 Jan '09

9:20 p.m.

Hi, What is the reason fusion::transform taskes a const Sequence? template< typename Sequence, typename F > typename result_of::transform::type transform( Sequence const& seq, F f); Does this means that the transform do not change the sequence itself o also its elements? Why the following do not compiles struct get { template<typename Sig> struct result; template<typename T> struct result { 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 t; boost::fusion::result_of::transform< boost::tuple , 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 = ??(boost::fusion::transform(t, get())); boost::as_vector? Is there a as_tuple function? Should this force the evaluation of the get function? Thanks, Vicente

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Joel de Guzman

6 Jan 6 Jan

10:26 p.m.

vicente.botet wrote:

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Hi,

What is the reason fusion::transform taskes a const Sequence? template< typename Sequence, typename F > typename result_of::transform::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.

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Why the following do not compiles struct get { template<typename Sig> struct result; template<typename T> struct result { 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 t; boost::fusion::result_of::transform< boost::tuple , 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 = ??(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 v; v = transform(v, f); Regards, -- Joel de Guzman http://www.boostpro.com http://spirit.sf.net

vicente.botet

11:24 p.m.

----- Original Message ----- From: "Joel de Guzman" To: Sent: Tuesday, January 06, 2009 11:26 PM Subject: Re: [Boost-users] [fusion] const and transform algorithm

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vicente.botet wrote:

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Hi,

What is the reason fusion::transform taskes a const Sequence? template< typename Sequence, typename F > typename result_of::transform::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? if yes, would the following get a transformation I'm locking for? vector,shared_pair > t; boost::fusion::for_each(t, intrusive_get()); vector res=unzip<1>(boost::fusion::for_each(t, intrusive_get())); where shared_pair would be a sequence something like a shared_ptr and unzip would be the reverse function of zip. BTW do you think that the counterpart function of zip has a place on Fusion?

...

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boost::tuple = ??(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 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? 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& 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? Thanks, Vicente

Joel de Guzman

7 Jan 7 Jan

12:58 a.m.

vicente.botet wrote:

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----- Original Message ----- From: "Joel de Guzman" To: 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::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

...
t; boost::fusion::for_each(t, intrusive_get()); vector res=unzip<1>(boost::fusion::for_each(t, intrusive_get()));

where shared_pair would be a sequence something like a shared_ptr 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?

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boost::tuple = ??(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 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.

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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.

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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

vicente.botet

6:29 a.m.

----- Original Message ----- From: "Joel de Guzman" To: Sent: Wednesday, January 07, 2009 1:58 AM Subject: Re: [Boost-users] [fusion] const and transform algorithm

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vicente.botet wrote:

...
----- Original Message ----- From: "Joel de Guzman" To: 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::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.

I just wat to apply the unction get to a tuple of futures and reviver the stored values. T& future<T>::get() is non const

...

...
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vector 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.

Thanks for the reference. I was looking for a function prototype. BTW, the example is not in line with the functions. vector v(12, 5.5f); std::cout << at_c<0>(v) << std::endl; std::cout << at_c<1>(v) << std::endl;

...

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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.

Great! Would you try with a Boost.Move library? Thanks, Vicente

Joel de Guzman

7:04 a.m.

vicente.botet wrote:

...

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

I just wat to apply the unction get to a tuple of futures and reviver the stored values. T& future<T>::get() is non const

YOu can use a tie (tuple of references).

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vector 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.

Thanks for the reference. I was looking for a function prototype. BTW, the example is not in line with the functions. vector v(12, 5.5f); std::cout << at_c<0>(v) << std::endl; std::cout << at_c<1>(v) << std::endl;

Why? "Semantics of an expression is defined only where it differs from, or is not defined in Random Access Sequence." You should read more carefully ;-)

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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.

Great! Would you try with a Boost.Move library?

Hah. Wish I had more time :-P Regards, -- Joel de Guzman http://www.boostpro.com http://spirit.sf.net

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vicente.botet