[Haskell-cafe] Flattening tail recursion?

[Georg Martius]

Hi Will,

you probably get confused with stack overflow through non-tail recursive function and stack overflow because you accumulate all intermediate values in the closure. It was allready posted before, that you need to enforce the evaluation of the + in order to get the function run in constant space. The thing is, that it is harder to achieve than I expected it to be.

countLines' ls = foldl (\x y -> let x' = x + 1 in x' `seq` y `seq` x' ) 0 ls

will run in constant space, but just if compiled with -O (ghc-6.2.1). The seq function forces the evaluation of its first argument (at least to Head Normal Form). The second one is just passed through. To be honest I don't understand why I need the optimisation option and why I do need to force the evaluation of y ?!. I find this really hard to figure out and I think the strictness analyser could be a bit more eager :-).

	Georg


On Fri, 10 Dec 2004 13:55:03 -0500, GoldPython <goldpython at gmail.com> wrote:

> I did this:
>
> countLines ls = foldl (\x y -> x + 1) 0 ls
>
> Still overflows.
>
>
> On Fri, 10 Dec 2004 19:07:04 +0100 (MEZ), Henning Thielemann
> <iakd0 at clusterf.urz.uni-halle.de> wrote:
>>
>>
>>
>> On Fri, 10 Dec 2004, Robert Dockins wrote:
>>
>> > >    countLines [] = 0
>> > >    countLines (_:ls) = 1 + countLines ls
>> > >
>> > > I would have thought that this was tail recursive and would be
>> > > flattened into iteration by the compiler. Can anyone explain why?
>> >
>> > countlines = aux 0
>> >     where aux x [] = x
>> >           aux x (_:ls)
>> >                 | x `seq` False = undefined
>> >                 | otherwise     = aux (x+1) ls
>> >
>> > The `seq` causes the x to be evaluated, so it should run in constant space.
>>
>> Is it also possible to do that with 'foldl'?
>>
>> Why is Prelude.length not defined this way (according to the Haskell98
>> report)?
>>
>>
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