Hi Greg,
But if I do something like this: f lst = show (filter (> 1) lst, filter (> 2) lst) then it looks like GHC won't garbage collect list elements until the first filter has completely finished
There is a very good reason for this, show (a,b) is essentially show (a,b) = "(" ++ show a ++ ", " ++ show b ++ ")" so all the elements of a are shown first, which means that lst is held up in a thunk of filter, to be evaluated later. In this particular case, you know that not (>2) implies not (>1) so you can speed it up with: f lst = show (part1, part2) where part1 = filter (>1) lst part2 = fitler (>2) part1 note that part2 only examines part1. As long as part1 is smaller than lst, this will reduce the space leak. There is really no way to "eliminate" the space leak in this circumstance, since you really do need to hold a part of the data in memory while you show the first one, so you can then show the second one. Thanks Neil