Re: [Haskell-cafe] instance of String illegal

29 Sep 2006

      Adam,
...
class Foo a where
    mkFoo :: a -> String
instance Foo String where
    mkFoo x = x
In addition to making use of language extensions or wrapper types,  
you could go with the following workaround in just plain Haskell 98:

   import List

   class MkFoo a where
     mkFoo     :: a -> String

     mkFooList :: [a] -> String
     mkFooList =  concat . intersperse ", " . map mkFoo

   instance MkFoo Char where
     mkFoo     = (: [])
     mkFooList = concatMap mkFoo

   instance (MkFoo a) => MkFoo [a] where
     mkFoo = mkFooList

For instance:
...
mkFoo False
   "no"
...
mkFoo [False, True]
   "no, yes"
...
mkFoo 'h'
   "h"
...
mkFoo "haskell"
   "haskell"
The same approach is taken for implementing Show in the standard  
libraries. Note that it requires you to fix the type constructors  
that are to be treated in a special manner ([] in the above example).

HTH,

   Stefan

Re: [Haskell-cafe] instance of String illegal

Stefan Holdermans