On Sun, Aug 3, 2008 at 11:06 AM, Arie Groeneveld <[EMAIL PROTECTED]> wrote:
> 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 >> >> >> >> > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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