Show Constraint type in haskell - haskell

Question

I am trying to use show function to print to the console value of zer or one, but I can not do it. Here is my code:
{-# LANGUAGE NoMonomorphismRestriction #-}
import Control.Arrow
import Data.List
import qualified Data.Map as M
import Data.Function
class Eq a => Bits a where
zer :: a
one :: a
instance Bits Int where
zer = 0
one = 1
instance Bits Bool where
zer = False
one = True
instance Bits Char where
zer = '0'
one = '1'
I am trying to use function show to convert zer or one to the string.
So I tried it:
k = zer
show k
but I got this error
<interactive>:10:1: error:
• Ambiguous type variable ‘a0’ arising from a use of ‘show’
prevents the constraint ‘(Show a0)’ from being solved.
Probable fix: use a type annotation to specify what ‘a0’ should be.
These potential instances exist:
instance (Show k, Show a) => Show (M.Map k a)
-- Defined in ‘containers-0.5.7.1:Data.Map.Base’
instance Show Ordering -- Defined in ‘GHC.Show’
instance Show Integer -- Defined in ‘GHC.Show’
...plus 24 others
...plus 11 instances involving out-of-scope types
(use -fprint-potential-instances to see them all)
• In the expression: show zer
In an equation for ‘it’: it = show zer
so i tried to create instance for show. So I added this to my code:
instance (Show a) => Show (Bits a) where
show zer = "0"
show one = "1"
But I got another error
main.hs:25:28: error:
• Expected a type, but ‘Bits a’ has kind ‘Constraint’
• In the first argument of ‘Show’, namely ‘Bits a’
In the instance declaration for ‘Show (Bits a)’
Can you tell me what I am doing wrong?

Answer 1

You're trying to make a class an instance of a class, rather than making a type an instance of a class. Compare:
Show a => Show (Bits a) -- Invalid
to
Show a => Show (Maybe a) -- Valid
where Maybe is a datatype whereas Bits is a class name.
I don't think it's possible to express "anything that has a Bits instance has a Show instance", because it can lead to overlapping instances: if you could define something like that, then when you use show :: Int -> String the compiler wouldn't know whether to use the Prelude's instance of Show Int or the show that would be defined by Int being an instance of Bits.
A messy workaround could be to enforce "the other direction": that every instance of Bits must be an instance of Show, which would allow you to use a's Show instance rather than your own one:
class (Show a, Eq a) => Bits a where
zer :: a
one :: a
main = print (zer :: Int)
although this requires an explicit type signature to resolve the ambiguity in the type of zer at the call site.