| Date: Wed, 28 Sep 2011 21:12:06 -0400 (EDT)
 | From: will@ccs.neu.edu
 |
 | ...
 |
 | I suspect that the
 | person(s) who ran the tests was unaware that the behavior of
 | NaNs is system-specific:  It depends on the hardware and the
 | numerical libraries as well as upon the parts of the system
 | that are under the control of an implementor of Scheme.

There is only one NaN in SCM.  So the behavior of NaNs in SCM is
independent of hardware and libraries:

  (eqv? +nan.0 (/ 0.0 0.0)) ==> #t

 | What's more, the result depends on the particular NaNs that
 | are involved in the test.  Here's an example from Larceny
 | running on a Macintosh:
 |
 | > (begin (define nan0 +nan.0)
 |          (define nan1 (/ 0.0 0.0))
 |          (define nan2 (- +inf.0 +inf.0)))
 |
 | > (eqv? nan0 nan0)
 | #t
 |
 | > (eqv? nan1 nan1)
 | #t
 |
 | > (eqv? nan2 nan2)
 | #t
 |
 | > (eqv? nan0 nan1)
 | #f
 |
 | > (eqv? nan0 nan2)
 | #f
 |
 | > (eqv? nan1 nan2)
 | #t

All of these return #t in SCM.

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