Exercise 1.5

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

Then we evaluates the expression

(test 0 (p))

What behavior will it with an interpreter that uses applicative-order evaluation?

(test 0 (p))
->(test 0 (p))
->(test 0 (p))
->.....

When the interpreter uses applicative-order evaluation, it evaluate first (p). And it will be evaluated as (p) all the time.
So, if the interpreter uses applicative-order, the 'test' procedure will not come to an end.

What behavior will it with an interpreter that uses normal-order evaluation?

(test 0 (p))
->(if (= 0 0) 0 (p))
->0

When the interpreter uses normal-order evaluation, the (p) procedure will be evaluated later.
So, the behavior will be an end.