Thank you very much! That solved it ;)I had to put explicit type signature in
front of advance in order to compile
From: cgae...@uwaterloo.ca
Date: Wed, 28 Nov 2012 08:01:38 -0500
Subject: Re: [Haskell-cafe] How to incrementally update list
To: edwards.b...@gmail.com
CC: bm...@hotmail.com; haskell-cafe@haskell.org
Here's a version that works:
import Control.DeepSeq
list = [1,2,3,4,5]
advance l = force $ map (\x -> x+1) l
run 0 s = srun n s = run (n-1) $ advance s
main = do let s = run 50000000 list putStrLn $ show s
The problem is that you build of a huge chain of updates to the list. If we
just "commit" each update as it happens, we'll use a constant amount of memory.
Haskell's laziness is tricky to understand coming from imperative languages,
but once you figure out its evaluation rules, you'll begin to see the elegance.
Ηope this helps,
- Clark
On Wed, Nov 28, 2012 at 7:07 AM, Benjamin Edwards <edwards.b...@gmail.com>
wrote:
TCO + strictnesses annotations should take care of your problem.
On 28 Nov 2012 11:44, "Branimir Maksimovic" <bm...@hotmail.com> wrote:
Problem is following short program:list = [1,2,3,4,5]
advance l = map (\x -> x+1) l
run 0 s = srun n s = run (n-1) $ advance s
main = do let s = run 50000000 list putStrLn $ show s
I want to incrementally update list lot of times, but don't knowhow to do this.
Since Haskell does not have loops I have to use recursion,but problem is that
recursive calls keep previous/state parameterleading to excessive stack.and
memory usage.I don't know how to tell Haskell not to keep previous
state rather to release so memory consumption becomesmanagable.
Is there some solution to this problem as I think it is rathercommon?
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