#| This exercise has no tests.
Any solution is a right answer. |#
(define (p) (p))
(define (test x y) 
  (if (= x 0) 
    0 
    y))

; (test 0 (p))
; test returns 0 when normal order is used 
; test goes into infinite recursion when
; applicative order evaluation is used 
; because applicative order evaluates its 
; argument before the body of procedure gets 
; executed and normal order doesn't evaluate 
; the arguments until they are actually needed.