http://www.haskell.org/onlinereport/exps.html#sect3.17
Arthur
On 3-mrt-05, at 15:39, Scott Turner wrote:
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.
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