[Haskell-beginners] Return type from class method

PATRICK BROWNE patrick.browne at dit.ie
Sun Oct 8 05:24:06 UTC 2017


Thomas,
Thanks for your response. I agree that using (+) in this manner leads to
name clashes.
My question is prompted by a research paper [1] where a form of lifting is
attempted  using  type classes.
I think that the original code in paper (below) is intended to be
illustrative rather than practical.
I have edited the code to compile and run (after a fashion, see below). But
I cannot get the functions to return the lifted type.
Also I have a problem compiling the lines with two lambdas, e.g. (np1 = xy
(\t -> 4.0 + 0.5 * t) (\t -> 4.0 - 0.5 * t))
Regards,
Pat
---------------------------------------------------------------------------------------------
Original code [1]:
------------------------------------------------------------------------------------------
class Number a where
(+), (-), (*) :: a -> a -> a
sqr, sqrt :: a -> a
sqr a = a * a

type Moving v = Time -> v

instance Number v => Number (Moving v) where
 (+) a b = \t -> (a t) + (b t)
 (-) a b = \t -> (a t) - (b t)
 (*) a b = \t -> (a t) * (b t)
 sqrt a = \t -> sqrt (a t)

class Number s => Points p s where
 x, y :: p s -> s
 xy :: s -> s -> p s
 dist :: p s -> p s -> s
 dist a b = sqrt (sqr ((x a) - (x b)) +
 sqr ((y a) - (y b)))

data Point f = Point f f

instance Number v => Points Point v where
 x (Point x1 y1) = x1
 y (Point x1 y1) = y1
 xy x1 y1 = Point x1 y1

instance Number v => (Point v) where
 (+) a b = xy (x a + x b) (y a + y b)
 (-) a b = xy (x a - x b) (y a - y b)


np1, np2 :: Point (Moving Float)
np1 = xy (\t -> 4.0 + 0.5 * t) (\t -> 4.0 - 0.5 * t)
np2 = xy (\t -> 0.0 + 1.0 * t) (\t -> 0.0 - 1.0 * t)
movingDist_1_2 = dist np1 np2
dist_at_1 = movingDist_1_2 1.0


----------------------------------------------------------------------------------------------
My attempt at getting above code to run:
-----------------------------------------------------------------------------------------------
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE TypeSynonymInstances #-}
module Moving where
data  Time  = Time Double
type Moving v  = Time -> v
data Point v  = Point v v deriving Show

class  Number a where
 (+),(-),(*)  ::  a -> a  -> a
 sqrt :: a -> a


instance (Fractional a,Floating a) => Number  (Moving a) where
 (+) a b = \t -> ((a t) Prelude.+ (b t))
 (-) a b = \t -> ((a t) Prelude.- (b t))
 (*) a b = \t -> ((a t) Prelude.* (b t))
 sqrt a =  \t -> Prelude.sqrt (a t)

a,b ::  Moving Double
a (Time x) = 4.0
b (Time x) = 4.0
testPlus ::(Moving Double)
testPlus = (a Moving.+ b)
testPlusArg = (a Moving.+ b) (Time 2.0)
testSqrt  = (Moving.sqrt a) (Time 2.0)



class (Number s) =>  Points p s  where
 x, y :: p s -> s
 xy :: s -> s -> p s
 dist :: p s -> p s -> s
 dist a b = Moving.sqrt (sqr ((x a) Moving.- (x b))  Moving.+ sqr ((y a)
Moving.- (y b)))
               where sqr z = z Moving.* z



-- instance (Floating v,Number v) => Points Point v  where
instance  (Number s) => Points Point s where
 x (Point x1 y1) = x1
 y (Point x1 y1) = y1
 xy x1 y1 = Point x1 y1

instance Number v =>  Number (Point v) where
 (+) a b = xy (x a Moving.+ x b) (y a Moving.+ y b)
 (-) a b = xy (x a Moving.- x b) (y a Moving.- y b)


md1,md2,md3,md4 ::  Moving Double
md1  (Time x) = 0.0
md2  (Time x) = 0.0
md3  (Time x) = 10.0
md4  (Time x) = 10.0
testMD1 = (md1  (Time 2.0))
testX =  x (Point md1 md2)  (Time 2.0)
testY =  y (Point md1 md2) (Time 2.0)
testXY =  (xy md1 md2)::(Point (Moving Double))
testX' =  x testXY  (Time 1.0)
testD = dist (Point md1 md2) (Point md3 md4)  (Time 1.0)

--- I cannot get the rest to work

[1] Ontology for Spatio-temporal Databases
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.113.9804&rep=rep1&type=pdf

On 7 October 2017 at 23:30, Thomas Jakway <tjakway at nyu.edu> wrote:

> You could hide Prelude and define it yourself in a different module but
> that would be a pretty bad idea.  Everyone who wanted to use it would have
> to import your module qualified and refer to it as MyModule.+, which
> defeats the point of making it (+) and not `myAdditionFunction` in the
> first place.
>
> Haskell deliberately doesn't allow overloading.  Having (+) return
> something other than Num would be extremely confusing.
>
> On 10/07/2017 05:07 AM, PATRICK BROWNE wrote:
>
> Hi,
> Is there a way rewriting the definition of (+) so that testPlusArg returns
> a (Moving Double). My current intuition is that the signature [(+)  ::  a
> -> a  -> a] says that the type should be the same as the arguments. And
> indeed (:t testPlus) confirms this. But the type of  testPlusArg is a
> Double.
>  Can I make it (Moving Double) ?
> Thanks,
> Pat
>
>
> {-# LANGUAGE FlexibleInstances #-}
> {-# LANGUAGE TypeSynonymInstances #-}
> module Moving where
> data  Time  = Time Double
> type Moving v  = Time -> v
>
> class  Number a where
>  (+)  ::  a -> a  -> a
>
> instance Number  (Moving Double) where
>  (+) a b = \t -> ((a t) Prelude.+ (b t))
>
> a,b ::  Moving Double
> a (Time x) = 2.0
> b (Time x) = 2.0
> testPlus ::(Moving Double)
> testPlus = (a Moving.+ b)
> testPlusArg = (a Moving.+ b) (Time 2.0)
>
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This email originated from DIT. If you received this email in error, please 
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Is ó ITBÁC a tháinig an ríomhphost seo. Má fuair tú an ríomhphost seo trí 
earráid, scrios de do chóras é le do thoil. Tabhair ar aird, mura tú an 
seolaí ainmnithe, go bhfuil dianchosc ar aon nochtadh, aon chóipeáil, aon 
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