Exercise 1.5.  Ben Bitdiddle has invented a test to determine whether the interpreterhe is faced with is using applicative-order evaluation or normal-orderevaluation. He defines the following two procedures:

(define (p) (p))



(define (test x y)

  (if (= x 0)

      0

      y))

Then he evaluates the expression

(test 0 (p))

What behavior will Ben observe with an interpreter that usesapplicative-order evaluation? What behavior will he observe with aninterpreter that uses normal-order evaluation?

Explain your answer.(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 evaluatethe consequent or the alternative expression.)

1、无參数函数

(define (p) (+ 1 2) )

;Value p

p

;Value 18:#[compound-procedure 18 p]

(p)

;Value 3

比較一下：

(define x (+ 1 2) )

;Value x

x

;Value 3

2、无限循环

(define (p) (p))无限循环。前一个(p)定义无參数函数，后一个(p)表示调用自己。

3.应用序和正则序的差别

应用序：(test 0 (p))时，先求值(p)——无限循环，再将(p)的值带入

(if (= x 0)

      0

      y))

正则序：(test 0 (p))时。先带入/展开，即

(if (= x 0)

      0

      (p)))

依照if求值顺序，推断0=0,结果为#t。

注意：if为正则序。