A follow-up question. I still haven't got the monadic version working, and the real use case involves IO actions. I looked at http://www.haskell.org/haskellwiki/Recursion_in_a_monad and adapted the 'tail-recursive' snippet on the page into main = do let f 0 acc = return acc f n acc = do v <- return 1 f (n-1) (v+acc) f 1000000 100 >>= print which still blows the memory. And so does this program main = do s<-newIORef (0::Int) mapM_ (\i->modifyIORef s (+1)) [0..10000000] readIORef s>>=print Why? ----- 原始邮件 ----- 发件人: sdiyazg@sjtu.edu.cn 收件人: haskell-cafe@haskell.org 发送时间: 星期四, 2012年 9 月 20日 上午 12:08:19 主题: Re: How to implement nested loops with tail recursion? Now I have discovered the right version... main = print (f 1 0::Int) where f i s = (if i<=20000 then (f (i+1) (s + g 1 0)) else s) where g j s = (if j<=20000 then (g (j+1) (s + i*j)) else s) ----- 原始邮件 ----- 发件人: sdiyazg@sjtu.edu.cn 收件人: haskell-cafe@haskell.org 发送时间: 星期三, 2012年 9 月 19日 下午 11:35:11 主题: How to implement nested loops with tail recursion? I need to implement fast two-level loops, and I am learning using seq to make calls tail-recursive. I write programs to compute main = print $ sum [i*j|i::Int<-[1..20000],j::Int<-[1..20000]] This program (compiled with -O2) runs twenty times slower than the unoptimized (otherwise the loop gets optimized out) C version. But it seems to run in constant memory, so I assume that it has been turned into loops. #include int main(){ int s=0; for(int i=1;i<=20000;++i){ for(int j=1;j<=20000;++j){ s+=i*j; } } printf("%d\n",s); return 0; } Then I write main = print $ f 1 where f i = let x = g 1 in x `seq` (x + if i<20000 then f (i+1) else 0) :: Int where g j = let x = i*j in x `seq` (x + if j<20000 then g (j+1) else 0) :: Int This version runs out of memory. When I scale the numbers down to 10000, the program does run correctly, and takes lots of memory. Even if I change the seqs into deepseqs, or use BangPatterns (f !i =... ; g !j = ...), the situation doesn't change. A monadic version import Control.Monad.ST.Strict import Control.Monad import Data.STRef.Strict main = print $ runST $ do s <- newSTRef (0::Int) let g !i !j = if (j<=10000) then modifySTRef s (+1)>>(g i (j+1)) else return () let f !i = if (i<=10000) then g i 1>>(f $ i+1) else return () f 1 readSTRef s also runs out of memory. So how can I write a program that executes nested loops efficiently?

So how do I force IO actions whose results are discarded (including IO ()) to be strict? main = do s<-newIORef (1::Int) let f :: Int -> Int -> IO Int f 0 !acc = return acc -- note strict accumulator f n !acc = do v <- modifyIORef s (+2) >>readIORef s -- reading immediately after writing f (n-1) (v+acc) f 1000000 100 >>= print readIORef s>>=print runs OK, while main = do s<-newIORef (1::Int) let f :: Int -> Int -> IO Int f 0 !acc = return acc -- note strict accumulator f n !acc = do v <- modifyIORef s (+2) >>return 1 f (n-1) (v+acc) f 1000000 100 >>= print readIORef s>>=print , main = do s<-newIORef (1::Int) let f :: Int -> Int -> IO Int f 0 !acc = return acc -- note strict accumulator f n !acc = do v <- modifyIORef s (+2) >>readIORef s>>return 1 f (n-1) (v+acc) f 1000000 100 >>= print readIORef s>>=print and main = do s<-newIORef (1::Int) let f :: Int -> Int -> IO Int f 0 !acc = return acc -- note strict accumulator f n !acc = do v <- (>>return 1) $! (modifyIORef s (+2) >>readIORef s) f (n-1) (v+acc) f 1000000 100 >>= print readIORef s>>=print all overflows after correctly printing the first number ----- 原始邮件 ----- 发件人: "Johan Tibell" 收件人: sdiyazg@sjtu.edu.cn 抄送: haskell-cafe@haskell.org 发送时间: 星期四, 2012年 9 月 20日 上午 1:28:47 主题: Re: [Haskell-cafe] How to implement nested loops with tail recursion? On Wed, Sep 19, 2012 at 7:24 PM, wrote:

main = do let f 0 acc = return acc f n acc = do v <- return 1 f (n-1) (v+acc) f 1000000 100 >>= print

Try this main = do let f :: Int -> Int -> IO Int f 0 !acc = return acc -- note strict accumulator f n acc = do v <- return 1 f (n-1) (v+acc) f 1000000 100 >>= print

On Wed, Sep 19, 2012 at 8:00 PM, wrote:

So how do I force IO actions whose results are discarded (including IO ()) to be strict?

In your particular case it looks like you want Data.IORef.modifyIORef'. If your version of GHC doesn't include it you can write it like so: -- |Strict version of 'modifyIORef' modifyIORef' :: IORef a -> (a -> a) -> IO () modifyIORef' ref f = do x <- readIORef ref let x' = f x x' `seq` writeIORef ref x' -- Johan