Are all arrows functors?
|
View:
New views
6 Messages
—
Rating Filter:
Alert me
|
 |
Are all arrows functors?
by Neil Brown-7
::
Rate this Message:
Hi,
I was thinking about some of my code today, and I realised that where I have an arrow in my code, A b c, the type (A b) is also a functor. The definition is (see http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html): fmap = (^<<) -- Or, in long form: fmap f x = arr f <<< x Out of curiosity, and since this is a typical haskell-cafe question, does this definition of fmap hold for all arrows? And is there a wiki page somewhere that has a table of all of these Haskell type-classes (Functor, Monad, Category, Arrow, Applicative and so on), and says that if you are an instance of class A you must have some corresponding instance of B? (e.g. all Monads are Functors and Applicatives) I'm fairly certain my arrow isn't a Monad or Applicative, although of course it must be a Category, given the type-class dependency, but it would be nice when using one of these things to see what other instances you should automatically supply. Thanks, Neil. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@... http://www.haskell.org/mailman/listinfo/haskell-cafe |
|
|
Re: Are all arrows functors?
by jkff
::
Rate this Message:
2009/11/3 Neil Brown <nccb2@...>:
> Hi, > > I was thinking about some of my code today, and I realised that where I have > an arrow in my code, A b c, the type (A b) is also a functor. The > definition is (see > http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html): > > fmap = (^<<) > -- Or, in long form: > fmap f x = arr f <<< x > > Out of curiosity, and since this is a typical haskell-cafe question, does > this definition of fmap hold for all arrows? > > And is there a wiki page somewhere that has a table of all of these Haskell > type-classes (Functor, Monad, Category, Arrow, Applicative and so on), and > says that if you are an instance of class A you must have some corresponding > instance of B? (e.g. all Monads are Functors and Applicatives) I'm fairly > certain my arrow isn't a Monad or Applicative, although of course it must be > a Category, given the type-class dependency, but it would be nice when using > one of these things to see what other instances you should automatically > supply. > What about the Typeclassopedia (http://haskell.org/sitewiki/images/8/85/TMR-Issue13.pdf)? > Thanks, > > Neil. > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@... > http://www.haskell.org/mailman/listinfo/haskell-cafe > -- Eugene Kirpichov Web IR developer, market.yandex.ru _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@... http://www.haskell.org/mailman/listinfo/haskell-cafe |
|
|
Re: Are all arrows functors?
by Nicolas Pouillard-2
::
Rate this Message:
Excerpts from Neil Brown's message of Tue Nov 03 13:45:42 +0100 2009:
> Hi, > > I was thinking about some of my code today, and I realised that where I > have an arrow in my code, A b c, the type (A b) is also a functor. The > definition is (see > http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html): > > fmap = (^<<) > -- Or, in long form: > fmap f x = arr f <<< x > > Out of curiosity, and since this is a typical haskell-cafe question, > does this definition of fmap hold for all arrows? Yes, as shown by the 'WrappedArrow' newtype: http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Applicative.html#t%3AWrappedMonad -- Nicolas Pouillard http://nicolaspouillard.fr _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@... http://www.haskell.org/mailman/listinfo/haskell-cafe |
|
|
Re: Are all arrows functors?
by porges-2
::
Rate this Message:
See also the paper "Idioms are oblivious, arrows are meticulous,
monads are promiscuous" [1]. Functors can be extended to give applicative functors (idioms) which can then be extended to arrows, and then monads. So all arrows are also (applicative) functors. [1]: http://homepages.inf.ed.ac.uk/wadler/papers/arrows-and-idioms/arrows-and-idioms.pdf _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@... http://www.haskell.org/mailman/listinfo/haskell-cafe |
|
|
Re: Are all arrows functors?
by Andrew Coppin
::
Rate this Message:
Nicolas Pouillard wrote:
> Excerpts from Neil Brown's message of Tue Nov 03 13:45:42 +0100 2009: > >> Hi, >> >> I was thinking about some of my code today, and I realised that where I >> have an arrow in my code, A b c, the type (A b) is also a functor. The >> definition is (see >> http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html): >> >> fmap = (^<<) >> -- Or, in long form: >> fmap f x = arr f <<< x >> >> Out of curiosity, and since this is a typical haskell-cafe question, >> does this definition of fmap hold for all arrows? >> > > Yes, as shown by the 'WrappedArrow' newtype: > > http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Applicative.html#t%3AWrappedMonad > While I don't wish to suggest that "all arrows are functors" is false, I think the argument "yes, because this library says so" is not too strong. Let us not forget, according to *the libraries*, Double is in Enum - which, I think you'll agree, is just weird... _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@... http://www.haskell.org/mailman/listinfo/haskell-cafe |
|
|
Re: Are all arrows functors?
by David Menendez-2
::
Rate this Message:
On Thu, Nov 5, 2009 at 4:34 PM, Andrew Coppin
<andrewcoppin@...> wrote: > Nicolas Pouillard wrote: >> >> Excerpts from Neil Brown's message of Tue Nov 03 13:45:42 +0100 2009: >> >>> >>> Hi, >>> >>> I was thinking about some of my code today, and I realised that where I >>> have an arrow in my code, A b c, the type (A b) is also a functor. The >>> definition is (see >>> http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Arrow.html): >>> >>> fmap = (^<<) >>> -- Or, in long form: >>> fmap f x = arr f <<< x >>> >>> Out of curiosity, and since this is a typical haskell-cafe question, does >>> this definition of fmap hold for all arrows? >>> >> >> Yes, as shown by the 'WrappedArrow' newtype: >> >> >> http://www.haskell.org/ghc/docs/latest/html/libraries/base/Control-Applicative.html#t%3AWrappedMonad >> > > While I don't wish to suggest that "all arrows are functors" is false, I > think the argument "yes, because this library says so" is not too strong. > Let us not forget, according to *the libraries*, Double is in Enum - which, > I think you'll agree, is just weird... It's fairly simple to prove the functor laws using the arrow laws. Among the nine laws for arrows are a >>> arr id = a a >>> arr f >>> arr g = a >>> arr (g . f) Using the definition fmap f a = a >>> arr f, it's pretty simple to prove the functor laws: fmap id = id fmap f . fmap g = fmap (f . g) -- Dave Menendez <dave@...> <http://www.eyrie.org/~zednenem/> _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@... http://www.haskell.org/mailman/listinfo/haskell-cafe |
| Free embeddable forum powered by Nabble | Forum Help |
