[Haskellcafe] show in monad
Damien Mattei
damien.mattei at gmail.com
Thu Feb 28 14:30:58 UTC 2019
even with a definition of show i can not use it in flatten:
import Control.Monad
import Data.Ratio
import Data.List (all)
import Debug.Trace
newtype Prob a = Prob { getProb :: [(a,Rational)] } deriving Show
instance Show a => Show (Prob a) where
show (Prob [(x,r)]) = ((show x) ++ " _ " ++ (show r))
instance Functor Prob where
fmap f (Prob xs) = trace " Functor Prob "
Prob $ map (\(x,p) > (f x,p)) xs
flatten :: Prob (Prob a) > Prob a
flatten (Prob xs) = trace (" flatten " ++ (show xs))
Prob $ concat $ map multAll xs
where multAll (Prob innerxs,p) = trace (" multAll p= " ++ (show p) ++ " ")
map (\(x,r) > (x,p*r)) innerxs
monade.hs:23:44: error:
• No instance for (Show a) arising from a use of ‘show’
Possible fix:
add (Show a) to the context of
the type signature for:
flatten :: forall a. Prob (Prob a) > Prob a
• In the second argument of ‘(++)’, namely ‘(show xs)’
In the first argument of ‘trace’, namely
‘(" flatten " ++ (show xs))’
In the expression: trace (" flatten " ++ (show xs)) Prob

23  flatten (Prob xs) = trace (" flatten " ++ (show xs))
 ^^^^^^^
Failed, no modules loaded.
it seems show i defined is not in the context of flatten???
damien
On Thu, Feb 28, 2019 at 12:57 PM Jos Kusiek <jos.kusiek at tudortmund.de>
wrote:
> You simply cannot do that. To be more precise, you cannot use show inside
> the bind operator on Prob (but you could use it in flatten). Deriving Show
> creates a Show instance which looks something like that:
>
> instance Show a => Show (Prob a) where ...
>
> This instance needs "a" to instanciate Show, so you can only use show with
> Prob types, where "a" is an instance of Show itself, e. g. Prob Int. Your
> flatten function does not guarantee that "a" is an instance of Show. The
> type says, any type for "a" will do it. You can easily restrict that with a
> class constraint:
>
> flatten :: Show a => Prob (Prob a) > Prob a
>
> But now you have a problem with the bind operator. You can no longer use
> flatten here. The bind operator for Prob has the following type:
>
> (>>=) :: Prob a > (a > Prob b) > Prob b
>
> There are no constraints here and you cannot add any constraints. The type
> is predefined by the Monad class. So it is not guaranteed, that this Prob
> type has a show function and you cannot guarantee it in any way. So you
> cannot use show on your first parameter type (Prob a) or your result type
> (Prob b) inside the bind or any function that is called by bind.
>
> On 28.02.19 11:00, Damien Mattei wrote:
>
> just for tracing the monad i have this :
>
> import Control.Monad
>
> import Data.Ratio
> import Data.List (all)
> import Debug.Trace
>
> newtype Prob a = Prob { getProb :: [(a,Rational)] } deriving Show
>
> instance Functor Prob where
> fmap f (Prob xs) = trace " Functor Prob "
> Prob $ map (\(x,p) > (f x,p)) xs
>
>
> t
>
>
> flatten :: Prob (Prob a) > Prob a
> flatten (Prob xs) = trace (" flatten " ++ show xs)
> Prob $ concat $ map multAll xs
> where multAll (Prob innerxs,p) = trace " multAll "
> map (\(x,r) > (x,p*r)) innerxs
>
>
> instance Applicative Prob where
> pure = trace " Applicative Prob return " return
> (<*>) = trace " Applicative Prob ap " ap
>
> instance Monad Prob where
> return x = trace " Monad Prob return "
> Prob [(x,1%1)]
> m >>= f = trace " Monad Prob >>= "
> flatten (fmap f m)
> fail _ = trace " Monad Prob fail "
> Prob []
>
>
> {
> instance Applicative Prob where
>
> pure a = Prob [(a,1%1)]
>
> Prob fs <*> Prob as = Prob [(f a,x*y)  (f,x) < fs, (a,y) < as]
>
>
> instance Monad Prob where
>
> Prob as >>= f = Prob [(b,x*y)  (a,x) < as, let Prob bs = f a, (b,y) <
> bs]
>
> }
>
>
>
> in this :
>
> flatten :: Prob (Prob a) > Prob a
> flatten (Prob xs) = trace (" flatten " ++ show xs)
> Prob $ concat $ map multAll xs
> where multAll (Prob innerxs,p) = trace " multAll "
> map (\(x,r) > (x,p*r)) innerxs
>
>
> i have this error:
>
> [1 of 1] Compiling Main ( monade.hs, interpreted )
>
> monade.hs:22:43: error:
> • No instance for (Show a) arising from a use of ‘show’
> Possible fix:
> add (Show a) to the context of
> the type signature for:
> flatten :: forall a. Prob (Prob a) > Prob a
> • In the second argument of ‘(++)’, namely ‘show xs’
> In the first argument of ‘trace’, namely ‘(" flatten " ++ show xs)’
> In the expression: trace (" flatten " ++ show xs) Prob
> 
> 22  flatten (Prob xs) = trace (" flatten " ++ show xs)
>  ^^^^^^^
> Failed, no modules loaded.
>
> how can i implement a show for xs ?
> regards,
> damien
>
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>
> 
> Dipl.Inf. Jos Kusiek
>
> Technische Universität Dortmund
> Fakultät 4  Informatik / Lehrstuhl 1  Logik in der Informatik
> OttoHahnStraße 12, Raum 3.020
> 44227 Dortmund
>
> Tel.: +49 231755 7523
>
>
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