Re: [Haskell-cafe] help diagnosing space leak with IORef/STRef, just incrementing a million times.
On 13-01-07 12:12 AM, Thomas Hartman wrote:
I have a space leak in a function that increments a number inside
IORef or STRef (either lazy or strict).
IORef and STRef operations do not automatically evaluate contents.
writeIORef r (x + 1) simply stores a pointer to the expression (thunk)
x + 1 into the mutable cell. readIORef just reports back a pointer.
modifyIORef just calls readIORef and writeIORef. No evaluation throughout.
modifyIORef incr where
incr !x = x + 1
does not make a difference because it is just writeIORef r (incr x)),
i.e., simply stores a pointer to the expression (thunk) incr x into
the mutable cell. The whole process doesn't even care about how many
bangs are in incr.
(It is illuminating to consider how const True (incr x) does not
evaluate x. A pointer to True and a pointer to incr x are passed to
const, then const throws away the latter without even looking. See also
const True undefined. One day, you will thank writeIORef r
undefined; I certainly did.)
Same for both Data.STRef.Strict and Data.STRef.Lazy. They do not mean
what you think. Here is what they mean:
Data.STRef.Strict means what Control.Monad.ST.Strict means
Data.STRef.Lazy means what Control.Monad.ST.Lazy means
Control.Monad.ST.Strict means that the following hangs:
x = head (runST list) where
list :: ST s [Bool]
list = do {xs - list; return (True : xs)}
Control.Monad.ST.Lazy means that the above terminates and gives the
answer True.
(Up to this point, same story for Control.Monad.State.Strict and
Control.Monad.State.Lazy.)
I still have not understood Control.Monad.ST.Lazy enough to articulate
its full semantics, but I have some more examples to show what it does:
http://hpaste.org/63925
By understanding what Lazy in Control.Monad.ST.Lazy means, you also
see what Strict does *not* mean.
In IO or Control.Monad.ST.Strict, use
let y = x+1 in y `seq` write[IO/ST]Ref r y
to expedite the evaluation of x+1. Using the same idea, you may write
your own modify[IO/ST]RefNOW to evaluate while updating.
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[Haskell-cafe] help diagnosing space leak with IORef/STRef, just incrementing a million times.
I have had issues like this with modifySTRef before. Try to make a strict version modifySTRef' If memory serves, something like this worked for me. modifySTRef' r f = do a - readSTRef r writeSTRef r $! f a ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] help diagnosing space leak with IORef/STRef, just incrementing a million times.
A similar use-case and same solution with IORefs:
http://hpaste.org/diff/80055/80058 Guess which one threw a
stackoverflow and which one ran indefinitely when given a few hundred
million lines of input.
On 7 January 2013 07:35, Albert Y. C. Lai tre...@vex.net wrote:
On 13-01-07 12:12 AM, Thomas Hartman wrote:
I have a space leak in a function that increments a number inside
IORef or STRef (either lazy or strict).
IORef and STRef operations do not automatically evaluate contents.
writeIORef r (x + 1) simply stores a pointer to the expression (thunk) x
+ 1 into the mutable cell. readIORef just reports back a pointer.
modifyIORef just calls readIORef and writeIORef. No evaluation throughout.
modifyIORef incr where
incr !x = x + 1
does not make a difference because it is just writeIORef r (incr x)),
i.e., simply stores a pointer to the expression (thunk) incr x into the
mutable cell. The whole process doesn't even care about how many bangs are
in incr.
(It is illuminating to consider how const True (incr x) does not evaluate
x. A pointer to True and a pointer to incr x are passed to const, then
const throws away the latter without even looking. See also const True
undefined. One day, you will thank writeIORef r undefined; I certainly
did.)
Same for both Data.STRef.Strict and Data.STRef.Lazy. They do not mean what
you think. Here is what they mean:
Data.STRef.Strict means what Control.Monad.ST.Strict means
Data.STRef.Lazy means what Control.Monad.ST.Lazy means
Control.Monad.ST.Strict means that the following hangs:
x = head (runST list) where
list :: ST s [Bool]
list = do {xs - list; return (True : xs)}
Control.Monad.ST.Lazy means that the above terminates and gives the answer
True.
(Up to this point, same story for Control.Monad.State.Strict and
Control.Monad.State.Lazy.)
I still have not understood Control.Monad.ST.Lazy enough to articulate its
full semantics, but I have some more examples to show what it does:
http://hpaste.org/63925
By understanding what Lazy in Control.Monad.ST.Lazy means, you also see
what Strict does *not* mean.
In IO or Control.Monad.ST.Strict, use
let y = x+1 in y `seq` write[IO/ST]Ref r y
to expedite the evaluation of x+1. Using the same idea, you may write your
own modify[IO/ST]RefNOW to evaluate while updating.
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