On Mon, Feb 14, 2005 at 02:31:54PM +0900, Shin-Cheng Mu wrote:
Malcolm Wallace <Malcolm.Wallace@cs.york.ac.uk> wrote:
Just a comment, since a couple of people have made similar statements. Haskell will derive Eq for arbitrarily complex types - there is no restriction to "simple" types, whatever they might be.
Now that this topic is brought up...
Occasionally I would need to define recursive datatypes using an explicit fixed-point operator, such as:
data Fix f = In (f (Fix f)) deriving (Show, Eq) data L a x = Nil | Cons a x deriving (Show, Eq)
However, Haskell was not able to derive from Fix f any instances.
[snip]
This is rather unsatisfactory, because I would not be able to inspect values of type Fix f in the interpreter. Is there a way to get around this?
sincerely, Shin
Funny this comes up at this time, as Fix was on-topic yesterday at #haskell. One way to make Fix an instance of Show/Eq is this (based on http://www.haskell.org/hawiki/PreludeExts, where Fix is called Rec): class RecShow f where recShow :: Show a => f a -> String instance (RecShow f) => Show (Rec f) where show (In x) = "(In (" ++ recShow x ++ "))" instance RecShow Maybe where recShow = show instance RecShow [] where recShow = show instance Show a => RecShow (Either a) where recShow = show Happy Hacking, Remi -- Nobody can be exactly like me. Even I have trouble doing it.