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.
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! Amanda -- Amanda Clare http://users.aber.ac.uk/ajc99/ Tel: +44 (0)1970 621787 Fax: +44 (0)1970 622455 Dept. of Computer Science, University of Wales, Aberystwyth, SY23 3DB _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
