Re: [Haskell-cafe] Flattening tail recursion?

Just compiled this with -O and it ran with no stack overflow. Evidently, no `seq` needed for this one. Using ghc 6.2.2. countLines l = countLines' 0 l countLines' n [] = n countLines' n (_:ls) = countLines' (n + 1) ls On Fri, 10 Dec 2004 20:32:07 +0100, Georg Martius wrote:

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 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 wrote:

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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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