Looks like that takes us back to the original proposal.
-Edward
On Sun, Aug 25, 2013 at 10:22 AM, Reid Barton
[Apologies to Henning and Edward for duplicate email]
On Sun, Aug 25, 2013 at 5:48 AM, Henning Thielemann < schlepptop@henning-thielemann.de> wrote:
Am 24.08.2013 19:52, schrieb Edward Kmett:
I would like to replace this instance with
instance a ~ Char => IsString [a] where fromString = id
Your complaint proves my concerns about those FlexibleInstances. The best instance is a Haskell 98 instance:
class IsCharList a where fromCharList :: [a] -> String
instance IsCharList Char where fromCharList = id
instance IsCharList a => IsString [a] where fromString = fromCharList
This is both the most flexible solution and it is portable.
http://www.haskell.org/**haskellwiki/List_instancehttp://www.haskell.org/haskellwiki/List_instance
This doesn't compile (in the IsString instance, we need fromString :: String -> [a], but we have fromCharList :: [a] -> String), and if you fix the declaration of fromCharList to fromCharList :: String -> [a], it fails to achieve the goal that was the original purpose of Edward's proposal. See below, where I've used RebindableSyntax to write a custom IsString class to avoid overlapping with the built-in instance IsString String.
{-# LANGUAGE RebindableSyntax, OverloadedStrings #-}
import Prelude(Char, String, id, length, print)
class IsString a where fromString :: String -> a
class IsCharList a where fromCharList :: String -> [a]
instance IsCharList Char where fromCharList = id
instance IsCharList a => IsString [a] where fromString = fromCharList
main = print (length "abc")
{- /tmp/is.hs:17:22: Ambiguous type variable `a0' in the constraint: (IsCharList a0) arising from the literal `"abc"' Probable fix: add a type signature that fixes these type variable(s) In the first argument of `length', namely `"abc"' In the first argument of `print', namely `(length "abc")' In the expression: print (length "abc") -}
Regards, Reid Barton
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