On Monday 11 February 2002 02:10, Hal Daume III wrote:
> So instaed of storing 1 and 2 in the list, it stores pointers to
> them. The reason it does this is so that it makes it easy to generate
> code for polymorhpic functions. (,) being boxed means that instead of
> storing a pair of element, y
Good luck! This is not for the
faint of heart.
J
| -Original Message-
| From: Jorge Adriano [mailto:[EMAIL PROTECTED]]
| Sent: Sunday, February 10, 2002 7:18 PM
| To: [EMAIL PROTECTED]
| Subject: Re: efficiency question
|
|
| On Sunday 10 February 2002 18:48, Kirsten Chevalier wrote:
|
I don't know of a reference, but here's the basic idea:
if you have a polymorphic type (like [a]), and it's instantiated to [Int]
(or whatever), instead of storing in memory a standard linked list
representation like
[1,2] = 1:2:[] =
[1 x] /--> [2 x] /--> Nil
\
On Sunday 10 February 2002 18:48, Kirsten Chevalier wrote:
> I'd guess that it's not just that you have to apply the (,) constructor --
> it also has to do with the fact that the tuples it's constructing here are
> boxed.
could you elaborate a little more on that (boxed / unboxed) or provide a li
On Sat, Feb 09, 2002 at 12:01:02PM -0500, [EMAIL PROTECTED]
wrote:
>
> > I'd say that's because in the second case you also got to apply the (,),
> > besides the (+)/(-) constructor during the transversing...
> > Am I right?
>
> opss... I meant to write: the (,) constructor besides the (+)/(-).
> I'd say that's because in the second case you also got to apply the (,),
> besides the (+)/(-) constructor during the transversing...
> Am I right?
opss... I meant to write: the (,) constructor besides the (+)/(-)...
J.A.
___
Glasgow-haskell-users ma
On Friday 08 February 2002 22:14, Hal Daume III wrote:
> define
>
> test1 l =
> let s1 = foldr (+) 1 l
> s2 = foldr (-) 1 l
> in (s1, s2)
>
> test2 l =
> let s = foldr (\x (a,b) -> (x+a,x-b)) (1,1) l
> in s
>
> why is test1 so much faster than test2 for long lists l (eg
>
define
test1 l =
let s1 = foldr (+) 1 l
s2 = foldr (-) 1 l
in (s1, s2)
test2 l =
let s = foldr (\x (a,b) -> (x+a,x-b)) (1,1) l
in s
why is test1 so much faster than test2 for long lists l (eg
[1..100])? replacing foldr with foldl makes it faster (of course),
but
