On Thu, Jan 29, 2009 at 12:44 AM, Eugene Kirpichov
With the Y combinator, the code becomes as follows:
f = \somedata1 somedata2 -> fst $ fix (\(aa,bb) -> (AA somedata1 bb, BB somedata2 aa))
I tried to prove to myself that the structure really turns out cyclic, but games with reallyUnsafePtrEquality# didn't lead to success; that's why it's "reallyUnsafe" :-/
Your definition with "fix" isn't lazy enough, so it goes into an infinite loop. You need to match lazily on the pair; here's a better body for "fix": fix (\p -> (AA somedata1 $ snd p, BB somedata2 $ fst p)) To prove that the structure really turns out cyclic, you can use Debug.Trace: import Debug.Trace (trace) import Data.Function (fix) data AA = AA Int BB deriving Show data BB = BB Int AA deriving Show f = \data1 data2 -> fst $ fix $ \p -> (trace "eval_aa" $ AA data1 $ snd p, trace "eval_bb" $ BB data2 $ fst p) sumAA 0 _ = 0 sumAA n (AA v bb) = trace "sumAA" (v + sumBB (n-1) bb) sumBB 0 _ = 0 sumBB n (BB v aa) = trace "sumBB" (v + sumAA (n-1) aa) main = print $ sumAA 10 $ f 1 2 *Main> main eval_aa sumAA eval_bb sumBB sumAA sumBB sumAA sumBB sumAA sumBB sumAA sumBB 15 -- ryan