[Haskell-beginners] Stack space overflow: using strict accumulator still fails
Hugo Ferreira
hmf at inescporto.pt
Thu Oct 27 15:22:26 CEST 2011
Hello,
After trying the suggestions, I still cannot execute
the code. I have tried:
scoreRule_ r zs = Z.foldlz' (scoreElem r) (0, 0) zs
where scoreElem r !s@(!nCorrect, !nIncorrect) z =
case ruleApplication r z of
Just tag -> if tag == correct then
(nCorrect+1, nIncorrect)
else (nCorrect, nIncorrect + 1)
Nothing -> s
where (correct, _) = Z.cursor z
and
scoreRule_ r zs = Z.foldlz' (scoreElem r) (0, 0) zs
nIncorrect)
where (correct, _) = Z.cursor z
nCorrect = nCorrect + 1
nIncorrect = nIncorrect + 1
where scoreElem r s@(nCorrect, nIncorrect) z =
case ruleApplication r z of
Just tag -> if tag == correct then
let nCorrect = nCorrect + 1 in
nCorrect `seq` nIncorrect `seq`
(nCorrect, nIncorrect)
else let nIncorrect = nIncorrect + 1 in
nCorrect `seq` nIncorrect `seq`
(nCorrect, nIncorrect)
Nothing -> s
where (correct, _) = Z.cursor z
In an attempt to figure out the problem I also tried:
scoreRule_ r zs = Z.foldlz' (scoreElem r) (0, 0) zs
where scoreElem r !s@(!nCorrect, !nIncorrect) z = (nCorrect,
nIncorrect)
where (correct, _) = Z.cursor z
nCorrect = nCorrect + 1
nIncorrect = nIncorrect + 1
Strangely enough GHC complains that correct, nCorrect and nIncorrect
are not used. Why is this so for nCorrect and nIncorrect? Why won't the
above also execute in a strict manner?
Does anyone have any ideas why I still get stack-overflow?
TIA,
Hugo F.
On 10/27/2011 12:41 PM, Daniel Fischer wrote:
> On Thursday 27 October 2011, 12:45:11, Hugo Ferreira wrote:
>> Hello,
>>
>> Have a stack overflow but cannot see why (read up on [1],
>> may be missing something trivial). Once again using the
>> http://nlpwp.org/ book code. If I call the following function,
>> it blows its top:
>>
>> scoreRule :: TransformationRule -> Z.Zipper (Tag, Tag) -> Int
>> scoreRule r z = nCorrect - nIncorrect
>> where (nCorrect, nIncorrect) = scoreRule_ r z
>>
>> scoreRule_ :: TransformationRule -> Z.Zipper (Tag, Tag) -> (Int, Int)
>> scoreRule_ r = Z.foldlz' (scoreElem r) (0, 0)
>> where scoreElem r s@(nCorrect, nIncorrect) z =
>> case ruleApplication r z of
>> Just tag -> if tag == correct then
>> (nCorrect + 1, nIncorrect)
>> else
>> (nCorrect, nIncorrect + 1)
>> Nothing -> s
>> where (correct, _) = Z.cursor z
>>
>> however I see that the eager version of foldlz is being used.
>> I also though that maybe ruleApplication my not be executing
>> immediately. But I cannot see why (added definition below for
>> reference).
>>
>> Can anyone point out why this is not strict?
>
> The additions (increments) are never forced before the final subtraction,
> so from scoreRule_ you will probably get a pair of thunks
> (((...(0+1)+1...)+1), ((...(0+1)+1...)+1)),
> since the forcing in the fold can only use seq, to force *the outermost
> constructor* of the intermediate results, in this case, the outermost
> constructor is the pair constructor - (,) - and the components are left
> unforced.
>
> To force the increments without (big) delay, you can
> - use a custom strict pair type instead of ordinary pairs
>
> data P = P !Int !Int -- {-# UNPACK #-} the fields for extra goodness
>
> so that forcing the outermost constructor automatically forces the
> components.
>
> - make the scoreElem function strict in the components of the accumulator
> s, with ghc {-# LANGUAGE BangPatterns #-},
>
> scoreElem r !s@(!nCorrect, !nIncorrect) z = ...
>
> that way you will never get bigger thungks than (n+1) in the components
>
> - force the updated count as it is constructed,
>
> if tag == correct
> then let newCorrect = nCorrect+1
> in newCorrect `seq` (newCorrect, nIncorrect)
> else let newIncorrect = ...
>
>
> The important thing to be aware of is that seq only forces the outermost
> level of a value. If the value is a structure with more levels, it doesn't
> prevent the building of huge thunks in the inner levels at all.
> You then have to take care of that yourself, by using a datatype with the
> desired strictness or, in the case of folds and similar, providing a
> comination function with the desired strictness.
>
> HTH,
> Daniel
>
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