1.5.txt

(define (p) (p))

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

(test 0 (p)) ;Test


Application: Applicative order evaluation evaluates the operand sub expressions
before evalutating the operator with the results. (p) evaluates to (p) so :

(test 0 (p))

reduces to

(test 0 (p))

which reduces to

(test 0 (p))

and so on.


(test 0 (p))
(test 0 (p))
(test 0 (p))

Compare the above to 
(define z (7 + 1))
using
(test 0 z)

which reduces to

(test 0 (7+1))
(test 0 8)
(if (= 0 0) 0)
(if true 0)
0

The first expression (test 0 (p)) using applicative ordering never allows its second
operand to fully evaluate, therefor cannot continue reducing.

Normal-order:
evaluates to 0 because (p) is never needed. Normal order evaluation only
evaluates operands when ABSOLUTELY needed.