;;;; A Vec is move-only: passing one to a function transfers ownership, and the
;;;; source binding is dead afterwards. That rule is what makes a double free
;;;; unrepresentable, which is why `free` needs no analysis of its own.
(defn take [v (Vec i32)] i32
  (let [n (len v)]
    (free v)
    n))

(defn main [] i32
  (let [v (vec-new i32)]
    (push v 1)
    (println (take v))
    ;; v went with the call. Being refused here is the whole test.
    (println (len v))
    0))