(let ((u (vector))
(v (vector)))
(list (list 'eq (eq? u v) (eq-hash u) (eq-hash v)
(= (eq-hash u) (eq-hash v)))
(list 'eqv (eqv? u v) (eqv-hash u) (eqv-hash v)
(= (eqv-hash u) (eqv-hash v)))))
;Value 15: ((eq #f 107689176 107689168 #f) (eqv #t 107689176 107689168 #f))
Oops.
Why do we treat empty vectors as eqv? If we do, eqv-hash needs to be
made to agree.
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