On Mon, Oct 11, 2010 at 00:51, Ozgur Akgun <ozgurak...@gmail.com> wrote: > My point was: you need to find/define two operators, not just one. That > still holds :)
No it doesn't. f $ g $ h $ x == f (g (h x)) == f . g . h $ x == x $$ h $$ g $$ f if you have the correct associativity for ($$) --Max _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
