On Thu, May 10, 2007 at 06:13:16PM +0100, Neil Mitchell wrote:
>
> >> Also remember that evaluating an expression in Haskell is _really_
> >> hard!
> >
> >Really? Looks pretty damn simple to me...
>
> In that case I throw down the challenge of writing an interpetter that
> takes a Haskell syntax tree and evaluates it :)
>
> What is the value of show [] ? Remember that show ([] :: String) /=
> show ([] :: [Bool), so I can't see how you can drop the types and keep
> the semantics.
show [] by itself can't be evaluated even with type inference, though. A
more convincing example, IMO, is something like
class Foo a where
foo :: a -> String
instance Foo Bool where
foo _ = "Bool"
instance Foo Char where
foo _ = "Char"
x :: String
x = foo (id $! (undefined :: Bool))
where to evaluate x you can't evaluate the argument to foo (in general
it might not terminate) in order to find out what type this instance of
foo needs to have..
Thanks
Ian
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