Much better. Although I struggle a bit with the exercises ;) Let's see
id :: a -> a curry :: ((a,b) -> c) -> a -> b -> c => curry id :: ((a,b) -> (a,b)) -> a -> b -> (a,b) So basically if f = curry id then f x y = (x,y) which means curry id = (,) something like this? Do I win a price now? ;) -----Original Message----- From: Henning Thielemann [mailto:[EMAIL PROTECTED] Sent: Tuesday, July 03, 2007 3:21 PM To: Peter Verswyvelen Cc: Haskell-Cafe@haskell.org Subject: Re: [Haskell-cafe] Haskell's currying and partial application On Tue, 3 Jul 2007, Peter Verswyvelen wrote: > IMHO when reading > > http://haskell.org/haskellwiki/Currying > > a newbie like me cannot see the difference between currying and partial application... Better now? _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
