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] ?
_______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
