Space leaks, and Writers, and Sums (oh my!)

08022021  
Вопрос
I've been playing with the Writer Monad recently, and I've run into what appears to be a space leak. I can't say I fully understand these things yet, so I'd like to know what's happening here, and how to fix it.
First, here's how I can trigger this error:
import Control.Monad.Writer
import Data.Monoid
foo :: Integer > Writer (Sum Integer) Integer
foo 0 = return 0
foo x = tell (Sum x) >> foo (pred x)
main = print $ runWriter $ foo 1000000
I get:
Stack space overflow: current size 8388608 bytes.
Use `+RTS Ksize RTS' to increase it.
To understand this better, I've reimplemented similar functionality without Writer or Sum, and if I keep things nice and lazy, I get the same error:
bar :: Integer > (Integer, Integer)
bar x = bar' 0 x
where bar' c 0 = (0, c)
bar' c x = bar' (c + x) (pred x)
But I can remedy this by adding seq
to the equation:
bar' c x = c `seq` bar' (c + x) (pred x)
I've tried seq
ing various bits of my foo
function, but that doesn't seem
to help. Also, I've tried using Control.Monad.Writer.Strict
but that
doesn't make a difference either.
Does Sum
need to be strict somehow? Or am I missing something
completely different?
Notes
 I may have my terminology wrong here. According to Space leak
zoo, my problem would be classified as a 'stack overflow', and if
that's the case, how would I convert
foo
to a more iterative style? Is my manual recursion the problem?  After reading Haskell Space Overflow, I had
the idea to compile with
O2
, just to see what happens. This may be a topic for another question, but with optimizations, even myseq
'dbar
function fails to run. Update: This issue goes away if I addfnofulllaziness
.
Решение
The problem with Writer monad is that it's >>=
is not tailrecursive:
instance (Monoid w, Monad m) => Monad (WriterT w m) where
m >>= k = WriterT $ do
(a, w) < runWriterT m
(b, w') < runWriterT (k a)
return (b, w `mappend` w')
As you can see it has to evaluate both m
and k a
to evaluate mappend
which means the entire stack of recursive calls has to be forced before the first mappend
can be evaluated. I believe that regardless of the strictness Writer monad will cause stack overflow in your definition (or can it be avoided with lazy version somehow?).
If you still want to use monads you can try State
which is tailrecursive. Either strict version of it with strict put
:
import Control.Monad.State.Strict
foo :: Integer > State (Sum Integer) Integer
foo 0 = return 0
foo x = do
w < get
put $! w `mappend` (Sum x)
foo (pred x)
main = print $ (`execState` Sum 0) $ foo 1000000
Or lazy version with continuation passing style (CPS):
import Control.Monad.Cont
import Control.Monad.State.Lazy
foo :: Integer > ContT Integer (State (Sum Integer)) Integer
foo 0 = return 0
foo x = do
w < get
put $! w `mappend` (Sum x)
foo (pred x)
main = print $ ((`execState`Sum 0) . (`runContT`return)) $ foo 1000000
Handy analog for tell
:
stell :: (MonadState w m, Monoid w) => w > m ()
stell a = get >>= \w > put $! w `mappend` a
I suspect that if it was possible to use ContT
with Writer
CPS would help us with Writer
as well, but looks like it isn't possible to define ContT for MonadWriter:
Другие советы
Look at the source for the strict writer monad: http://hackage.haskell.org/packages/archive/transformers/0.2.2.0/doc/html/src/ControlMonadTransWriterStrict.html#line122
The difference from the lazy writer is that in the latter, the pattern matches are lazy  but in neither case is the mappend
operation or the "state" of the writer thus far forced. To solve your issue, you'd need a "superstrict" writer.
I think your understanding is correct.
I'm interested in how these functions:
bar :: Integer > (Integer, Integer)
bar x = bar' 0 x
where bar' c 0 = (0, c)
bar' c x = c `seq` bar' (c + x) (pred x)
 bar' c x = let c' = c+x in c' `seq` bar' c' (pred x)
 bar' !c !x = bar' (c+x) (pred x)
produces a stack overflow when compiled with optimizations, although the related functions:
bar2 :: Integer > (Integer, Integer)
bar2 x = bar' 0 x
where bar' c 0 = (0, c)
bar' !c !x = let c' = c+x in c' `seq` bar' c' (pred x)
bar3 :: Integer > Integer
bar3 x = bar' 0 x
where bar' c 0 = c
bar' c x = c `seq` bar' (c + x) (pred x)
bar4 :: Integer > (Integer, Integer)
bar4 x = (0, bar' 0 x)
where bar' c 0 = c
bar' c x = c `seq` bar' (c + x) (pred x)
do not.
I think this looks like a bug in GHC's optimizer, and you should report it. From looking at the core (produced with ddumpsimpl
), the c
argument isn't strictly evaluated in all cases for the overflowing functions. bar2
is the closest working version I found to the original, and it seems overspecified to me.
Since you have a very simple test case, there's a good chance it'll get fixed.