# producing and consuming lists

**Jorge Adriano
**
jadrian@mat.uc.pt

*Tue, 5 Nov 2002 16:04:58 +0000*

I might have been not very clear in my last mail. I decided to post again=
, and=20
go straight to the point, with some small examples.
Consider the following function streams.
streams :: (Int->Bool, Int->Bool)->(Int, Int)->([Int],[Int])
streams (p,q) (x,y) =3D (xs',ys')
where
(xs,ys) =3D streams (p,q) ((x+y),(y-x))
xs' =3D if p x then x:xs else xs
ys' =3D if q y then y:xs else ys
- produces a pair of ('infinite') lists
- produced lists are not indepentent (you need to calculate elements of o=
ne=20
list to calculate elements of the other)
- in each recursive call an element is added to the 1st/2nd list iff it=20
satisfies a given (Int->Bool) function p/q
How should one consume (part of) both lists, avoiding space leaks?
A particular example of consuming both lists might be writing them to fil=
es:
main :: IO()
main =3D do
let (s1,s2)=3Dstream ... -- stream applied to some arguments (p,q)=
(x,y)
p' =3D ...=20
q' =3D ...=20
writeFile "f1.txt" (show$ takeWhile p' s1)
writeFile "f2.txt" (show$ takeWhile q' s2)
In this example all elements of s2 required to evaluate (takeWhile p' s1)=
are=20
kept in memory, until the first file is writen. Notice that writing one=20
element from s1 and one from s2 successively might still cause space leak=
s to=20
arise. Fusing the consuming functions with the producer is a possible, bu=
t=20
IMO dirty, way out.=20
If my question doesn't seem to make sense for any reason, please tell me,=
=20
maybe I am missing something obvious here.=20
Thanks,
J.A.