Alan Watson wrote:
> I know of at least two R5RS
> implementations in which the following need not evaluate to #t:
It gets worse.
I also know of lots and lots of R5RS implementations in which the
following WILL NOT evaluate to #t:
(let* ((x 1.0)
(foo (lambda (y) (eq? x y)))
(obj1 x)
(obj2 (+ 0.0 x)))
(eqv? (foo obj1) (foo obj2)))
Normally, all that's needed here is that the implementation use boxed
inexacts and doesn't share them on the fly.
Regards,
Alan
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