module Main where import Control.Monad import Control.Concurrent class Processor p where ready :: p b c → [b → c] instance Processor (→) where ready = repeat --instance Processor [b → c] where --ready = id newtype FunList b c = FunList [b → c] instance Processor FunList where ready (FunList fl) = fl broadcast :: Processor p ⇒ p b c → [b] → [c] broadcast p bs = bcast ps bs [] where ps = ready p bcast [] _ cs = cs bcast _ [] cs = cs bcast ps bs cs = let (cp,nps) = rotate [] ps (cb,nbs) = rotate [] bs in bcast nps nbs (cp cb:cs) rotate :: [a] → [a] → (a,[a]) rotate os (x:[]) = (x,os) rotate os (x:xs) = (x,xs) main :: IO () main = do let pid = id let ppm = FunList [ (x +) | x ← [1..10] ] print $ broadcast ppm [1..10] ------ Thank you for your answer! This comes close to solving the problem but in the last line of the above I want to be able to say: either

print $ broadcast id [1..10]

or

print $ broadcast [ (x +) | x ← [1..10] ] [1..10]

both need to be possible*. So is there a way to make the FunList disappear completely? Regards, Cetin P.S.: * broadcast is a dummy function, I need this for tidying up the interface of a little experiment: http://corsis.blogspot.com/ 2009/2/13 Benedikt Huber

Cetin Sert schrieb:

...

Hi,

class Processor a where ready :: (forall b c. a → [b → c])

instance Processor (b → c) where ready = repeat ... ------------------------------- Why can I not declare the above instances and always get: Hi Cetin, in your class declaration you state that a (Processor T) provides a function ready :: T -> [b -> c] so ready (t::T) has type (forall b c. [b -> c]), a list of functions from arbitrary types b to c.

The error messages tell you that e.g.

...

repeat (f :: t1 -> t2) has type (t1->t2) -> [t1->t2] and not the required type (t1->t2) -> [a -> b]

With your declarations,

...

head (ready negate) "hi" has to typecheck, that's probably not what you want.

...

Is there a way around this?

Maybe you meant

...

class Processor a where ready :: a b c -> [b -> c] instance Processor (->) where ready = repeat newtype FunList b c = FunList [b->c] instance Processor FunList where ready (FunList fl) = fl

I think the newtype FunList is neccessary here. benedikt

...

message.hs:229:10: Couldn't match expected type `b' against inferred type `b1' `b' is a rigid type variable bound by the instance declaration at message.hs:228:20 `b1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:19 Expected type: b -> c Inferred type: b1 -> c1 In the expression: repeat In the definition of `ready': ready = repeat

message.hs:229:10: Couldn't match expected type `c' against inferred type `c1' `c' is a rigid type variable bound by the instance declaration at message.hs:228:24 `c1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:21 Expected type: b -> c Inferred type: b1 -> c1 In the expression: repeat In the definition of `ready': ready = repeat

message.hs:232:10: Couldn't match expected type `b1' against inferred type `b' `b1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:19 `b' is a rigid type variable bound by the instance declaration at message.hs:231:20 Expected type: [b1 -> c] Inferred type: [b -> c1] In the expression: id In the definition of `ready': ready = id

message.hs:232:10: Couldn't match expected type `c1' against inferred type `c' `c1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:21 `c' is a rigid type variable bound by the instance declaration at message.hs:231:24 Expected type: [b -> c1] Inferred type: [b1 -> c] In the expression: id In the definition of `ready': ready = id

Is there a way around this?

Regards, CS

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Thank you Benedikt! Thanks to your help I also figured out the way to do it using type families yesterday: -------- class Pro p where type I p type O p re :: p → [I p → O p] instance Pro (b → c) where type I (b → c) = b type O (b → c) = c re = repeat instance Pro [b → c] where type I [b → c] = b type O [b → c] = c re = cycle broadcast :: Pro p ⇒ p → [I p] → [O p] ... -------- Regards, Cetin 2009/2/13 Benedikt Huber

Cetin Sert schrieb:

...

Thank you for your answer!

This comes close to solving the problem but in the last line of the above I want to be able to say:

either

...

print $ broadcast id [1..10]

or

...

print $ broadcast [ (x +) | x ← [1..10] ] [1..10]

both need to be possible*.

So is there a way to make the FunList disappear completely? Hi Cetin, yes, if you're willing to use multi-parameter typeclasses: class Processor p b c | p -> b c where ready :: p -> [b -> c] instance Processor (b -> c) b c where ready = repeat instance Processor [b -> c] b c where ready = id broadcast :: Processor p b c => p -> [b] -> [c]

Maybe there are other possibilities as well. -- benedikt

...

Regards, Cetin

P.S.: * broadcast is a dummy function, I need this for tidying up the interface of a little experiment: http://corsis.blogspot.com/

2009/2/13 Benedikt Huber mailto:benjovi@gmx.net>

Cetin Sert schrieb: > Hi, > > class Processor a where > ready :: (forall b c. a → [b → c]) > > instance Processor (b → c) where > ready = repeat > ... > ------------------------------- > Why can I not declare the above instances and always get: Hi Cetin, in your class declaration you state that a (Processor T) provides a function > ready :: T -> [b -> c] so > ready (t::T) has type (forall b c. [b -> c]), a list of functions from arbitrary types b to c.

The error messages tell you that e.g. > repeat (f :: t1 -> t2) has type > (t1->t2) -> [t1->t2] and not the required type > (t1->t2) -> [a -> b]

With your declarations, > head (ready negate) "hi" has to typecheck, that's probably not what you want.

> Is there a way around this?

Maybe you meant

> class Processor a where > ready :: a b c -> [b -> c] > instance Processor (->) where > ready = repeat > newtype FunList b c = FunList [b->c] > instance Processor FunList where > ready (FunList fl) = fl

I think the newtype FunList is neccessary here. benedikt

> > message.hs:229:10: > Couldn't match expected type `b' against inferred type `b1' > `b' is a rigid type variable bound by > the instance declaration at message.hs:228:20 > `b1' is a rigid type variable bound by > the type signature for `ready' at message.hs:226:19 > Expected type: b -> c > Inferred type: b1 -> c1 > In the expression: repeat > In the definition of `ready': ready = repeat > > message.hs:229:10: > Couldn't match expected type `c' against inferred type `c1' > `c' is a rigid type variable bound by > the instance declaration at message.hs:228:24 > `c1' is a rigid type variable bound by > the type signature for `ready' at message.hs:226:21 > Expected type: b -> c > Inferred type: b1 -> c1 > In the expression: repeat > In the definition of `ready': ready = repeat > > message.hs:232:10: > Couldn't match expected type `b1' against inferred type `b' > `b1' is a rigid type variable bound by > the type signature for `ready' at message.hs:226:19 > `b' is a rigid type variable bound by > the instance declaration at message.hs:231:20 > Expected type: [b1 -> c] > Inferred type: [b -> c1] > In the expression: id > In the definition of `ready': ready = id > > message.hs:232:10: > Couldn't match expected type `c1' against inferred type `c' > `c1' is a rigid type variable bound by > the type signature for `ready' at message.hs:226:21 > `c' is a rigid type variable bound by > the instance declaration at message.hs:231:24 > Expected type: [b -> c1] > Inferred type: [b1 -> c] > In the expression: id > In the definition of `ready': ready = id > > Is there a way around this? > > Regards, > CS > > >

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