> > > cpsfold f a [] ú > > > cpsfold f a (x:xs) ÿ x a (\y -> cpsfold f y xs) > > > > and f takes a continuation, Bob's my uncle, and I have a program that > > runs quickly in constant space! > > Good. I'm curious to know from other readers whether > continuations like this are the only way of solving it, > though.
Actually, and quite apart from it being cumbersome to use, I've got my doubts about whether this cpsfold really does the job (is that just me missing some point?-). Also, I'm curious to know why the usual strict variant of foldl doesn't help in this case? foldl' f a [] = a foldl' f a (x:xs) = (foldl' f $! f a x) xs or, with the recently suggested idiom for strictness, tracing and other annotations:-) annotation = undefined strict a = seq a False foldl' f a l | strict a = annotation foldl' f a [] = a foldl' f a (x:xs) = foldl' f (f a x) xs Claus _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
