gary ng wrote:

Still a bit confused.

My function is simply

f x y = if x < 100 then x + y else x

Try this: sum_to_100 xs = foldr (\x k a -> if a < 100 then k $! (x+a) else a) id xs 0 As expected, (sum_to_100 [1..]) gives 105, without trying to find the end of an infinite list. You get your early-out semantics combined with strict evaluation. Udo.

On Fri, 30 Sep 2005, gary ng wrote:

...

This should work as expected: takeWhile (

Thanks. But how would I think about using scanl instead of foldl(or foldl') when I want is the sum, but not the progressive result. Once again show me that I need to throw away all imperative stuff.

No problem: last (takeWhile (

Again, it depends how takeWhile is implemented -- if it's not tail recursive, the compiler will usually manage to run such functions in constant space. Bob On 30 Sep 2005, at 16:02, gary ng wrote:

Once again, many thanks to all who taught me about this small little problem. Don't even know there is init/last and thought there is only head/tail.

But just for my curiosity, would the takeWhile still store the intermediate result till my result is reached ? If so, and my list is really very long(and I need to go to 1/2 of its length), I would still use a lot more memory than imperative method or even the foldl one(where in both case, I just take one element) ?

--- Henning Thielemann wrote:

...

No problem: last (takeWhile (

The first sum which exceeds the limit could be computed with head (dropWhile (<=maxX) (scanl (+) 0 xs))

__________________________________ Yahoo! Mail - PC Magazine Editors' Choice 2005 http://mail.yahoo.com _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Am Freitag, 30. September 2005 17:14 schrieb Henning Thielemann:

On Fri, 30 Sep 2005, gary ng wrote:

...

Once again, many thanks to all who taught me about this small little problem. Don't even know there is init/last and thought there is only head/tail.

http://www.haskell.org/ghc/docs/6.4/html/libraries/base/Data.List.html

...

But just for my curiosity, would the takeWhile still store the intermediate result till my result is reached ? If so, and my list is really very long(and I need to go to 1/2 of its length), I would still use a lot more memory than imperative method or even the foldl one(where in both case, I just take one element) ?

If you don't trust the compiler you can at least test if the rest of the list is ignored: Run with an infinite list.

E.g. last (takeWhile (<10) (scanl (+) 0 (repeat 1)))

I think Gary wanted to know whether the initial part of scanl's result is stored. I think, it shouldn't, because of the 'last'. However, when profiling your versions versus testFoldl :: (a -> Bool) -> (a -> b -> a) -> a -> [b] -> a testFoldl _ _ z [] = z testFoldl p _ z _ | p z = z testFoldl p f z (x:xs) = testFoldl p f (f z x) xs, I found that your head (dropWhile ... ) took about twice as long and allocated roughly 2.6 times as much memory as mine and last (takeWhile ... ) was a bit worse. Still, it didn't use near enough memory to store takeWhile (<= 5*10^7) (scanl (+) 0 (repeat 1)), all three used (if I interpret the profiling graphs [-hc, -hb] correctly) about 16.5 k of memory for practically the complete runtime. Besides, head (dropWhile (<10) (scanl (+) 0 (replicate 9 1))) will raise an error, as will last (takeWhile ...) if the starting value satisfies the break-condition. Cheers, Daniel