Why do continuations not have useful arity?

Question

Consider the following code:

(call-with-values
    (lambda ()
      (call/cc (lambda (k)
         (k k k))))
  (lambda (x y)
    (procedure-arity y)))

It's pretty obvious here that the continuation at the point of the call/cc call is the lambda on the right-hand side, so its arity should be 2. However, the return value of the above (in Racket) is (arity-at-least 0) instead.

Indeed, running similar code in Guile (substituting procedure-minimum-arity for procedure-arity) shows that the continuation also supposedly allows any number of arguments, even though it's pretty clearly not the case.

So, why is that? As far as I understand (correct me if my understanding is wrong), the arity of a continuation is pretty straightforward: it's 1 except in the context of call-with-values, in which case it's whatever the arity of the right-hand-side lambda is. (Which, granted, can be complicated if it's a case-lambda or the like, but no more complicated than if you were calling (procedure-arity (case-lambda ...)) directly.)

Answer 1

A simpler way to see the same is:

(call-with-values
  (lambda () (error 'arity "~v" (procedure-arity (call/cc (λ (k) k)))))
  (lambda (x y) (procedure-arity y)))

and even simpler:

(procedure-arity (call/cc (λ (x) x)))

And for your question -- it's clear in the first case that the continuation expects two inputs, but cases like that are not too common. Eg, they're usually such examples, whereas "real code" would use define-values or have some unknown continuation, where the continuations that call/cc creates can have different arities depending on the context that they were created at. This means that there's not much point in trying to figure out these rare cases where the continuation's arity is known.

Footnote:

;; nonsensical, but shows the point
(define (foo) (call/cc (λ (x) x)))
(define x (foo))
(define-values [y z] (foo))