Is the behavior of evaluating z unspecified?
  z = f (0, z)
  f x = case x of
 (1,1) -> z
 _ -> 0
Hugs and GHC agree that z evaluates to 0. However, if the first line is 
changed to
  z = f (z,0)
then both implementations loop. In other words, the behavior depends on order 
of evaluation, which AFAIK is not specified.
_______________________________________________
Haskell mailing list
Haskell@haskell.org
http://www.haskell.org/mailman/listinfo/haskell

Reply via email to