On Sat, 22 Dec 2007, Cristian Baboi wrote: > On Sat, 22 Dec 2007 17:13:55 +0200, Philippa Cowderoy <[EMAIL PROTECTED]> > wrote: > > > Here's a trivial example that does so: > > > > (\x -> x) (\x -> x) > > > > A lambda calculus classic that doesn't typecheck in Haskell: > > > > (\x -> x x) (\x -> x x) > > > Feel free to try evaluating it! > > Thank you for your message. > > I tryed and this is what I've got: > ERROR - cannot find "show" function for: > *** Expression : (\x -> x) (\x -> x) > *** Of type : a -> a >
Yep, that's because while it can evaluate it down to (\x -> x) your interpreter doesn't know how to print the result. You can demonstrate that it works by then passing in something to that result though: ((\x ->x) (\x -> x)) 1 You'll have to evaluate the other one by hand. Don't spend too long with it though! -- [EMAIL PROTECTED] "The reason for this is simple yet profound. Equations of the form x = x are completely useless. All interesting equations are of the form x = y." -- John C. Baez _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
