[Haskell-cafe] IORef memory leak

[Daniel van den Eijkel]

Don Stewart schrieb:
> dvde:
>   
>> Don Stewart schrieb:
>>     
>>> It is not possible to write a modifyIORef that *doesn't* leak memory!
>>>   
>>>       
>> Why? Or can one read about it somewhere?
>>     
>
>
> Try writing a version of this program, using modifyIORef only, 
> such that it doesn't exhaust the heap:
>
>     import Data.IORef
>     import Control.Monad
>     import System.IO.Unsafe
>
>     ref :: IORef Int
>     ref = unsafePerformIO $ newIORef 0
>     {-# NOINLINE ref #-}
>
>     main = do
>         modifyIORef ref (\a -> a + 1)
>         main
>
> Run it in a constrained environment, so you don't thrash:
>
>     $ ./A +RTS -M100M
>     Heap exhausted;
>     Current maximum heap size is 99999744 bytes (95 MB);
>     use `+RTS -M<size>' to increase it.
>
> The goal is to run in constant space.
>
> -- Don
>
>   
Hm, do you say it is not possible to write a modifyIORef function that 
does not leak memory, or do you say it is not possible to use the 
(existing) modifyIORef without having memory leaks?

I wrote the following which runs in constant space, but it introduces 
strictness, which is not always desirable. And yes, using only 
modifyIORef this could not be done this way, because the strict 
evaluation happens on the IO-Monad-level. But such examples occured 
already in this thread.

import Data.IORef
import Control.Monad
import System.IO.Unsafe

ref :: IORef Int
ref = unsafePerformIO $ newIORef 0
{-# NOINLINE ref #-}

main = do
    myModifyIORef ref (\a -> a + 1)
    main
   
myModifyIORef :: IORef a -> (a->a) -> IO ()
myModifyIORef ref f = do
 a <- readIORef ref
 let a' = f a
 seq a' $ writeIORef ref a'

So would it make sense to create a strict modifyIORef' function?

best regards,
daniel