(define w (id (id 10)));;; L-Eval input:
count
;;; L-Eval value:
1
because when define w, call (id (id 10)), parameter (id 10) is delayed. so id only call once. count = 1.
;;; L-Eval input:
w
;;; L-Eval value:
10
when enter w in prompt, call (actual-value w),(id 10) is evaluated, id is called once more. so now w = 10, count = 2.
;;; L-Eval input:
count
;;;; L-Eval value:
2
meteorgan