seek for some explanation on SICP exercise 1.5

Question

The question can be found here.

In the book, I found one description of normal order evaluation was:

\"An alternative evaluation model would not evaluate the

Answer 1

You're right that an applicative-order evaluator would not terminate if asked to evaluate (p). However, the question at hand mentions that if has a particular evaluation semantics which is shared by the applicative and normal-order evaluators. Specifically, "Assume that the evaluation rule for the special form if is the same whether the interpreter is using normal or applicative order: The predicate expression is evaluated first, and the result determines whether to evaluate the consequent or the alternative expression."

The code from the exercise is

(define (p) (p))

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

and the test under consideration is

(test 0 (p))

Normal-order evaluation is the "fully expand and then reduce" option. Under normal-order evaluation, (test 0 (p)) is fully expanded as

(test 0 (p)) ==
(if (= 0 0)
  0
  (p))

Since if has the semantics described above, and the test condition in the expansion is (= 0 0), which is true, the normal-order evaluator determines to evaluate the consequent, which is 0, so the value of the expression is 0.

Using the applicative-order evaluation however, the first step in evaluating (test 0 (p)) is to evaluate the expressions test, 0, and (p), and then call ^{("apply", hence "applicative")} the value of test with the values produced by evaluating 0 and (p). Since the evaluation of (p) will not complete, neither will the evaluation of (test 0 (p)).