1) Yes, you are correct.

2) Yes, that is fine.  Sometimes people will use operators that don't conflict with prelude such as (+:), (-:) and (*:) to avoid these clashes.

On Tue, Oct 3, 2017 at 3:15 PM, PATRICK BROWNE <patrick.browne@dit.ie> wrote:
David,
Thank you for you informative and helpful reply.
I think that are two issues impeding my understanding of the original code.
1. The use of lambda
2. The structure of the class and instance

1. The use of lambda
Is seems that the arguments to (+!) , below ,must be functions. I was trying to use values
data Time  = Time Double 

type Moving v  = Time -> v

(+!) a b = \t -> (a t)  Prelude.+ (b t)

b ::  Moving Double

b (Time x) = x * 1.5

test = (b +! b) (Time 10.0)

Is my take on this correct?


2. The structure of the class and instance.
Recall the original code:

type Moving v = Time -> v

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

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)

I believe that this would have to be changed to avoid a clash with the Prelude definitions.
Is the following structuring reasonable?

module MovingPoint where
data  Time  = Time Double
type Moving v  = Time -> v

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

instance (Floating v) => Number (Moving v) 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)
 sqr a =  \t -> (a t)  Prelude.* (a t)
 sqrt a =  \t -> Prelude.sqrt (a t)

b ::  Moving Double
b (Time x) = x Prelude.* 1.5
test = (b MovingPoint.+ b) (Time 10.0)

Thanks,
Pat






On 3 October 2017 at 15:01, David McBride <toad3k@gmail.com> wrote:
You can get some intuition for how this works by replacing "Moving v" with its definition "Time -> v".  Let's look at the + operation.

class Number a where
  (+) :: a -> a -> a
instance Number v => Number (Moving v)
instance Number v => Number (Time -> v)
(+) :: Number v => (Time -> v) -> (Time -> v) -> (Time -> v)

So each argument of + must take a Time, the end result must also take a Time, and whatever each argument returns must be a Number (and thus has + defined for it).  So you can sort of see how it works.  + for a Moving v takes a time, then passes that time to each of its arguments, then adds the result.

(+) a b = \t -> (a t)  Prelude.+ (b t)

data Time = Time Double -- For example.

Then you can make formulas that are rooted in time.  For example (contrived) if you are throwing a ball, the distance of the ball from you at time f could be something like the following:

balldistance ::  Moving Double
balldistance (Time f) = f * 1.2

ball1 :: Moving Double
ball1 = balldistance

ball2 :: Moving Double
ball2 = balldistance

-- the combined distance of both balls at time f
bothballs :: Moving Double
bothballs = ball1 + ball2

Then you can get the combined distance of both balls after 12 seconds, for example.

test :: Double
test = bothballs (Time 12.0)


    


On Tue, Oct 3, 2017 at 9:07 AM, PATRICK BROWNE <patrick.browne@dit.ie> wrote:
Hi,
I am trying to compile, run, and understand the following code from [1].

type Moving v = Time -> v

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

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)

I followed the compiler advice to produce the following version which compiles:

{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE TypeSynonymInstances #-}
module MovingPoint where
type Time  = Float -- Type synonym assumed, could it be data type??
type Moving v  = Time -> v

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

instance (Floating v) => Number (Moving v) 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)
 sqr a =  \t -> (a t)  Prelude.* (a t)
 sqrt a =  \t -> Prelude.sqrt (a t)

I do not know how to invoke any of the operations. In general I do know how to execute lambdas.
I do not understand the bracketed pairs e.g. (a t).
Any help on understanding and running the program would be appreciated.
Thanks,
Pat


[1] Ontology for Spatio-temporal Databases

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