Fri, 8 Feb 2002 17:24:40 -0800
(moved to haskell-caf=E9)
I ran Hal's code on my computer, and with test2 I get a stack overflow =
(so I had to use +RTS option for it to finish). test1 does not overflow =
stack (of standard size, I mean without +RTS). Which implies that test2 =
uses more stack space than test1. why would it use more stack if not =
because of laziness?
> -----Original Message-----
> From: Hal Daume III [mailto:hdaume@ISI.EDU]=20
> Sent: Friday, February 08, 2002 4:35 PM
> To: Jorge Adriano
> Cc: Konst Sushenko; email@example.com
> Subject: Re: efficiency question
> I agree that it's the overhead of (,), but I don't see why=20
> there would be
> any overhead for doing this.
> Hal Daume III
> "Computer science is no more about computers | firstname.lastname@example.org
> than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume
> On Sat, 9 Feb 2002, Jorge Adriano wrote:
> > On Friday 08 February 2002 23:52, Hal Daume III wrote:
> > > I've tried using a strict fold:
> > >
> > > foldl' f a  =3D a
> > > foldl' f a (x:xs) =3D (foldl' f $! f a x) xs
> > >
> > > but that has no effect (or minimal effect).
> > That wouldn't work even if if laziness is the problem=20
> because that would only=20
> > cause the elements of the list to be evaluated to head=20
> normal form, the=20
> > elements of the pair would not be evaluated so you'd have a=20
> 'suspension of =20
> > (minus and plus) operations'.
> > instead of=20
> > > (\x (a,b) -> (x+a,x-b))
> > try=20
> > > (\x (a,b) -> (((,) $! x-a)$! x-b) )
> > I just noticed that you were the one who sent me the DeepSeq module.
> > This is the kind of place where I want to use it.
> > Instead of $!, try $!!.
> > And Konst Sushenko wrote:
> > >>My guess is that it is due to the laziness of the=20
> > >>in (,)
> > Seems to me like lazyness is not the right guess because=20
> both functions Hall=20
> > first posted were lazy. So I think it's just the overhead=20
> of applying (,)=20
> > besides (+) and (-) in each step. Do I make sense or am I=20
> missing something?
> > J.A.