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 <[EMAIL PROTECTED]> 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 <[EMAIL PROTECTED]> 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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