On Sun, Aug 3, 2008 at 11:06 AM, Arie Groeneveld
Sorry, should go the forum.
Ok, thanks. In this case the list consists of 6-digit alphanumeric codes. So doing something like:
foldl1 (\x y -> g y) xs
No, that still doesn't force elements. Let's say g is (+1): f = \x y -> (+1) y foldl1 f [1,2,3] (1 `f` 2) `f` 3 (+1) 3 4 So we don't need to compute (+1) on any numbers but 3. The most direct way is to force the elements of the list: import Control.Parallel.Strategies seqList rwhnf (map g xs) Note that the notion of "compute" in this example is to WHNF, so for example if g produces lists, it will only evaluate far enough to determine whether the list is a nil or a cons, not the whole thing.
will do the job?
=@@i
Bulat Ziganshin schreef:
Hello Arie,
Sunday, August 3, 2008, 1:56:43 PM, you wrote:
*Main>> last . f $ xs
this way you will get only "spin" of list computed, not elements itself. something like sum should be used instead
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