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. Best wishes, Wolfgang _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
