On Fri, 28 Dec 2007 18:27:29 +0200, Achim Schneider <[EMAIL PROTECTED]> wrote:
Wolfgang Jeltsch <[EMAIL PROTECTED]> wrote:
Am Freitag, 28. Dezember 2007 07:49 schrieben Sie:
> On Thu, 27 Dec 2007 18:19:47 +0200, Wolfgang Jeltsch
> <[EMAIL PROTECTED]> wrote:
> > Am Donnerstag, 27. Dezember 2007 16:34 schrieb Cristian Baboi:
> >> I'll have to trust you, because I cannot test it.
> >>
> >> let x=(1:x); y=(1:y) in x==y .
> >>
> >> I also cannot test this:
> >>
> >> let x=(1:x); y=1:1:y in x==y
> >
> > In these examples, x and y denote the same value but the result
> > of x == y is _|_ (undefined) in both cases. So (==) is not
> > really equality in Haskell but a kind of weak equality: If x
> > doesn’t equal y, x == y is False, but if x equals y, x == y might
> > be True or undefined.
>
> Thank you.
>
> I can only notice that y always has an even number of 1, which is
> not the case for x :-)
Both have an infinite number of 1. Why do you say “always”? It
seems that you think of x and y as “variables” whose values change
over time. This is not the case. They both have a single value for
all time: the infinite list consisting only of 1.
Does that then mean that
[1] = [1,1]
No. If you try to learn Haskell don't listen to me. :-)
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
