Re: slight differences between non-termination and failure

From: Sebastian Fischer <sebf_at_informatik.uni-kiel.de>
Date: Wed, 02 Feb 2011 18:14:47 +0900

I have simplified the example further:

repeat1 x = x:y:ys
 where
  y:ys = repeat1 x

repeat2 x = extend (repeat2 x)
 where
  extend (y:ys) = x:y:ys

repeat3 x = extend (repeat3 x)
 where
  extend ys = x:ys

Now,

    null (repeat1 1)
    null (repeat3 1)

terminate while

    null (repeat2 1)

does not.

Sebastian



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Received on Wed Feb 02 2011 - 11:42:22 CET

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