Re: Bug in documentation for eq? ?

[David Kastrup <dak@gnu.org>]

Andy Wingo <wingo@pobox.com> writes:

> On Wed 20 Jun 2012 12:40, David Kastrup <dak@gnu.org> writes:
>
>>      Numbers and characters are not equal to any other object, but the
>>      problem is they're not necessarily `eq?' to themselves either.
>>      This is even so when the number comes directly from a variable,
>>
>>           (let ((n (+ 2 3)))
>>             (eq? n n))       => *unspecified*
>
> Note that this example is taken from R5RS section 6.1.
>
>> A variable reference can't really be anything except eq? to itself in
>> my opinion.
>
> Depends on inlining.  Numbers are not considered to have identity, so
> they may be copied in some situations.

I can't see this being such a situation.  The number 5 as such does not
have identity.  But each individual instance of the number 5 is a Scheme
object, and Scheme objects have identity.  That different instances of 5
may or may not compare eq?: no question about that.  But the same?
That's just silly.

> In summary, I think the documentation is correct.

I think it is completely absurd.  It would mean, for example, that
(memq x (list x))
is generally unspecified.  It would mean that things like
(eq? (car x) (car x))
are generally unspecified even when x is a pair.

We have
scheme@(guile-user)> (eq? +nan.0 +nan.0)
$8 = #f
scheme@(guile-user)> (eqv? +nan.0 +nan.0)
$9 = #t
scheme@(guile-user)> (= +nan.0 +nan.0)
$10 = #f
scheme@(guile-user)> (let ((x +nan.0)) (eq? x x))
$11 = #t
scheme@(guile-user)> (let ((x +nan.0)) (eqv? x x))
$12 = #t
scheme@(guile-user)> (let ((x +nan.0)) (= x x))
$13 = #f

And that makes sense since eqv? is supposed to apply to a superset of
eq? while = is working on numerical values.

Which of the above would you consider unspecified?

-- 
David Kastrup