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