Exercise 1.5

sicp-1-5.lisp

(define (p) (p)) 
(define (test x y) 
  (if (= x 0) 
      0 
      y)) 

applicative-order evaluation
   evaluate arguments and then apply   

normal-order evaluation
   fully expand and then reduce
(test 0 (p))
(if (= x 0) 0 y)
(if (= 0 0) 0 (p))

* To apply a compound procedure to arguments, evaluate the body of the procedure with each formal parameter replaced by the corresponding argument. 

I am having difficulty understanding how to evaluate (p).  However, under normal-order evaluation, it would be evaluated last (during the reduction).  With applicative-order evaluation, (p) would be evaluate first and applied to (test 0 results-of-(p)).  So (p) evaluates to (p) which evaluates to (p)...  infinite loop under applicative-order evaluation.  Under normal-order evaluation, the 0 = 0 condition would be met and (p) would not be evaluated therefore no infinite loop.