Andy Wingo <wi...@pobox.com> writes: > On Wed 20 Jun 2012 12:40, David Kastrup <d...@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
