On Tue, 3 Jul 2007, Peter Verswyvelen wrote: > 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?
You got it! > Do I win a price now? ;) I hoped that nobody would care about the puzzle, thus I have not thought about a price so far. :-] _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
