> Thanks for all the advice. In the end, I couldn't make $! work for me > (it always seems to be harder than I think it will be to use it, and $! > and deepSeq makes my code run slowly).
:-( > But a continuation passing style foldl worked wonderfully. As Jay Cox pointed out by email, my answer was rot because I had confused foldl and foldr > I now have: > > > cpsfold f a [] = a > > cpsfold f a (x:xs) = f 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. Jón -- Jón Fairbairn [EMAIL PROTECTED] 31 Chalmers Road [EMAIL PROTECTED] Cambridge CB1 3SZ +44 1223 570179 (after 14:00 only, please!) _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
