>As you've apparently discovered, the trick is to be lazy but not too
>lazy. That is, you want to generate the list lazily but compute a
>partial result (i.e., the running total of that part of the list
>processed so far) strictly.
Thanks for all reactions. Now my simplified examples
indeed run in constant space. Unfortunatelly, my original code
still suffer from the same problem and even '-O2' does not help.
>More importantly, understand how foldl' works and be ready to
>apply the same analysis and fix to any similar function.
This is what I cannot do for the moment. How do I find out what is
really going on. Any pointers to a relevant
articles/literature? As an example, here is my earlier attempt
to define strictFoldl1:
strictFoldl1' f (a:b:c) = let strictFoldl1' f a (b:c) =
let fab = seq a $ seq b $ f a b
in case c of
[] -> fab
otherwise -> seq fab (strictFoldl1' f fab c)
in strictFoldl1' f a (b:c)
However, it does not seem to do the trick, why Alastair's code does it:
foldl' :: (a -> b -> a) -> a -> [b] -> a
foldl' f a [] = a
foldl' f a (x:xs) = (foldl' f $! f a x) xs
foldl1' f (x:xs) = foldl' f x xs
Thanks.
Jan
--
-------------------------------------------------------------------------
Jan Kybic <[EMAIL PROTECTED]> Odyssee, INRIA, Sophia-Antipolis, France
or <[EMAIL PROTECTED]>,tel. work +33 492 38 7589, fax 7845
http://www-sop.inria.fr/robotvis/personnel/Jan.Kybic/
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