`seq` forces evaluation of only the top-level construct in its first argument. (($!) similarly for its second argument.) I would guess your "newcounts" are structured (probably a tuple or list), in which case you are not forcing evaluation deeply enough. See http://haskell.org/pipermail/haskell/2001-August/001581.html and followup article.
Dean Herington Amanda Clare wrote: > I have some code which is being unnecessarily lazy (and occupying too > much heap space). The code should read and process several files one by > one. What's happening is that all files get read in but the processing > is delayed by laziness, and the files are being retained. It looks > something like this (after simplification): > > main = do > result <- foldM countAFile initialcounts fileNameList > prettyprint result > > countAFile oldcounts filename = > do > compiledFile <- readAndCompile filename > let newcounts = countAssociations compiledFile oldcounts > return newcounts > > I would love to have used CPS, but this seems to force me to lift a huge > amount of code into both the IO monad and the CPS, as countAssociations > just calls another function, which calls another, and so on. > > If I change countAFile to use a redundant test to force evaluation of > newcounts: > > countAFile oldcounts filename = > do > compiledFile <- readAndCompile filename > let newcounts = countAssociations compiledFile oldcounts > if newcounts == newcounts > then return newcounts > else error "suitable error message" > > then the program runs beautifully (if a little slowly), occupying a > constant amount of heap. > > If instead I use a strict data type to store my counts, or use seq, then > I get no improvement, and it's just as if I hadn't used them. I > understand why a strict data type won't help me, as it's the function > calls that are lazy. But what about $! and seq? > > countAFile oldcounts filename = > do > compiledFile <- readAndCompile filename > let newcounts = countAssociations compiledFile oldcounts > return $! newcounts > > Does anyone understand why $! (or equivalently "seq") have no effect > here? Is there a better way to do this? > > Amanda > -- > Amanda Clare http://users.aber.ac.uk/ajc99/ > Tel: +44 (0)1970 621922 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 _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
