How does garbage collection work in an example like the one below? You memoize a function with some sort of lookup table, which stores function arguments as keys and function results as values. As long as the function remains in scope, the keys in the lookup table remain in memory, which means that the keys themselves always remain reachable and they cannot be garbage collected. Right?
So what do you do in the case where you know that, after some period of time, some entries in the lookup table will never be accessed? That is, there are no references to the keys for some entries remaining, except for the references in the lookup table itself. You'd like to allow the memory occupied by the keys to be garbage collected. Otherwise, if the function stays around for a long time, the size of the lookup table always grows. How do you avoid the space leak? I notice that there is a function in Data.IORef, mkWeakIORef :: IORef a -> IO () -> IO (Weak (IORef a)) which looks promising. In the code below, however, there's only one IORef, so either the entire table gets garbage collected or none of it does. I've been reading the paper "Stretching the storage manager: weak pointers and stable names in Haskell," which seems to answer my question. When I attempt to run the memoization code in the paper on the simple fib example, I find that -- apparently due to lazy evaluation -- no new entries are entered into the lookup table, and therefore no lookups are ever successful! So apparently there is some interaction between lazy evaluation and garbage collection that I don't understand. My head hurts. Is it necessary to make the table lookup operation strict? Or is it something entirely different that I am missing? -Rod On Thu, 10 Sep 2009 18:33:47 -0700 Ryan Ingram <ryani.s...@gmail.com> wrote: > > memoIO :: Ord a => (a -> b) -> IO (a -> IO b) > memoIO f = do > cache <- newIORef M.empty > return $ \x -> do > m <- readIORef cache > case M.lookup x m of > Just y -> return y > Nothing -> do let res = f x > writeIORef cache $ M.insert x res m > return res > > memo :: Ord a => (a -> b) -> (a -> b) > memo f = unsafePerformIO $ do > fmemo <- memoIO f > return (unsafePerformIO . fmemo) > > I don't think there is any valid transformation that breaks this, > since the compiler can't lift anything through unsafePerformIO. Am I > mistaken? > > -- ryan _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe