On Thu, Nov 23, 2006 at 02:46:59PM -0600, Alan Watson wrote:
>
> If inexacts are unboxed and +nan.0 has only one representation, then the
> implementor can most simply implement eqv? on inexacts with a bitwise
> comparison. In this case, you get #t.
>
> If inexacts are boxed, then the implementor can most simply implement
> eqv? on inexacts with a numerical comparison. In this case, you get #f.
I'm not sure I see what boxing has to do with it; I was thinking more:
If inexacts are currently living in floating-point registers, the
simplest way to compare them may be with a numerical comparison
instruction that considers them not equal. So, you get #f.
In a similar situation on different hardware, it may be that a bit
comparison is the preferred option. So, you get #t.
--
(let ((C call-with-current-continuation)) (apply (lambda (x y) (x y)) (map
((lambda (r) ((C C) (lambda (s) (r (lambda l (apply (s s) l)))))) (lambda
(f) (lambda (l) (if (null? l) C (lambda (k) (display (car l)) ((f (cdr l))
(C k))))))) '((#\J #\d #\D #\v #\s) (#\e #\space #\a #\i #\newline)))))
Received on Fri Nov 24 2006 - 03:59:23 UTC