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