In addition to what Hal said, I think that even assuming that you want to write a function like that that is only supposed to be applied to finite lists (like, say, prelude's 'and' function, which does not terminate on an infinite list of True values) you still cannot do it because haskell lists are not like C lists. There is no notion, to my understanding, of list circularity. There are finite and infinite lists. Your example of a list that you call a circular list is just a definition of a recursive function that produces an infinite list. So what you are almost asking is how to write a function that checks if a list has repeated sub-sequences, which is not of course precisely the same that you asked (because even a finite list can have repeated sub-sequences), but close. And this is just one of many problems taken care of by various string search algorithms. konst

-----Original Message----- From: Hal Daume III [mailto:hdaume@ISI.EDU] Sent: Wednesday, April 24, 2002 12:38 PM To: klusacek@atrey.karlin.mff.cuni.cz Cc: haskell-cafe@haskell.org Subject: Re: lazy comparison for equality ?

I don't think you can write such a function. For instance, how would you know whether [1..] is circular or not? In order to know that it's not you'd need to evaluate it fully.

-- Hal Daume III

"Computer science is no more about computers | hdaume@isi.edu than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume

On Wed, 24 Apr 2002 klusacek@atrey.karlin.mff.cuni.cz wrote:

...

Hi - I'm a Haskell beginner and I have a problem.

Let's have a list which may be normal list list1 = [1,2,3] or a circular list list2 = 1:2:list2

Now I'd like to have a function which tells me whether the given list is circular or not. This doesn't work:

circ l = l l circ2 l [] = False circ2 l (_:as) | l==as = True | True = (circ2 l as)

It seems that comparison l==as really compares element by

...

falling into an infinite loop. I would need to compare

element thus pointers instead of

...

values.

Does anybody know how this could be done ?

Thanks.

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