Maybe not related, but does the following prove next is called once and only
once.
import qualified Data.ByteString as BS
import qualified Data.ByteString.Char8 as BSC
next =
do
nextcache - BS.readFile next.cache
let nextint = readInt (BSC.unpack nextcache)
Sorry for the last mail, I now tried it and it returns the next value every
time I call it.
I was using an unsafeperformIO trick somewhere, and that fas the one
resulting in the previously described behaviour.
You can just ignore the previous mail.
2009/12/17 Ozgur Akgun ozgurak...@gmail.com
Subject: Re: [Haskell-cafe] Haskell and memoization
To: Daniel Fischer daniel.is.fisc...@web.de, Gregory Crosswhite
gcr...@phys.washington.edu
Cc: haskell-cafe@haskell.org
Date: Wednesday, December 16, 2009, 12:58 AM
Hi all,
I think this (#3 below) is where I got the idea:
http://en.wikipedia.org
Am Mittwoch 16 Dezember 2009 15:49:54 schrieb michael rice:
Thanks all,
OK, so this definition of fib
fib 0 = 1
fib 1 = 1
fib n = fib (n-1) + fib (n-2)
would involve a lot of recomputation for some large n,
Where large can start as low as 20; 60 would be out of reach.
which memoization
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/Memoization
Since (I've read) Haskell never computes the value
of a function more than once, I don't understand the
need for memoization.
Enlighten me.
Michael
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://www.haskell.org/haskellwiki/Memoization
Since (I've read) Haskell never computes the value
of a function more than once, I don't understand the
need for memoization.
Enlighten me.
Michael
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Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite:
Haskell does not maintain a cache mapping function calls to their values,
so if you have some function f and call it with, say, the argument 7 in two
different places in your code, then it will re-evaluate the function at
each
Hmm, you raise an
On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote:
Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite:
Not even then, necessarily. And it's not always a good idea.
f k = [1 .. 20^k]
You raise a really good point here. One can force sharing, as I
return q, as occurs in memoization.
Michael
--- On Tue, 12/15/09, Gregory Crosswhite gcr...@phys.washington.edu wrote:
From: Gregory Crosswhite gcr...@phys.washington.edu
Subject: Re: [Haskell-cafe] Haskell and memoization
To: Daniel Fischer daniel.is.fisc...@web.de
Cc: haskell-cafe@haskell.org
Am Mittwoch 16 Dezember 2009 05:47:20 schrieb Gregory Crosswhite:
On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote:
Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite:
Not even then, necessarily. And it's not always a good idea.
f k = [1 .. 20^k]
You raise a really good
...@phys.washington.edu wrote:
From: Gregory Crosswhite gcr...@phys.washington.edu
Subject: Re: [Haskell-cafe] Haskell and memoization
To: Daniel Fischer daniel.is.fisc...@web.de
Cc: haskell-cafe@haskell.org
Date: Tuesday, December 15, 2009, 11:47 PM
Hmm, you raise an
On Dec 15, 2009, at 8:28 PM, Daniel
Am Mittwoch 16 Dezember 2009 07:22:42 schrieb Gregory Crosswhite:
#3 is true for Haskell, it's just that when your function call appears in
two different places, it is counted as two different expressions. Each
separate expression will only be evaluated once, though. This is what is
really
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