Re: [Haskell-cafe] show in monad

it's not clear, redaing back the thread it seems that it is not possible in a monad with the bind operator to have a display with show and now that adding the constraint and changing the type of flatten will do it, so i change the definition of flatten but it does not compile: flatten :: Show a => 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 Prelude> :load monade.hs [1 of 1] Compiling Main ( monade.hs, interpreted ) monade.hs:37:13: error: • No instance for (Show b) arising from a use of ‘flatten’ Possible fix: add (Show b) to the context of the type signature for: (>>=) :: forall a b. Prob a -> (a -> Prob b) -> Prob b • In the second argument of ‘trace’, namely ‘flatten’ In the expression: trace " Monad Prob >>= " flatten (fmap f m) In an equation for ‘>>=’: m >>= f = trace " Monad Prob >>= " flatten (fmap f m) | 37 | flatten (fmap f m) | ^^^^^^^ Failed, no modules loaded. Damien On Thu, Feb 28, 2019 at 4:58 PM Jos Kusiek wrote:

You do not need to change the Show instance. The one generated by deriving Show is fine. As I said, you need to change the type of flatten and add the constraint.

flatten :: Show a => Prob (Prob a) -> Prob a

On 28.02.19 15:30, Damien Mattei wrote:

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 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 Otto-Hahn-Straße 12, Raum 3.020 44227 Dortmund

Tel.: +49 231-755 7523

-- Dipl.-Inf. Jos Kusiek

Technische Universität Dortmund Fakultät 4 - Informatik / Lehrstuhl 1 - Logik in der Informatik Otto-Hahn-Straße 12, Raum 3.020 44227 Dortmund

Tel.: +49 231-755 7523