SICP习题 1.6 (再探 函数 与 正则序 应用序 关系)

<一>

(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))

(new-if (= 2 3) 0 5)
(new-if (= 1 1) 0 5)

new-if在两种环境下均运行正常.

(if (= 2 3) 0 5)
(if (= 1 1) 0 5)

if 在R5RS下输出正常,在Lazy Racket下为(delay ...)

<二>

(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))

(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x)
x)))

(define (improve guess x)
(average guess (/ x guess)))

(define (average x y)
(/ (+ x y) 2))

(define (good-enough? guess x)
(< (abs (- (square guess) x)) 0.001))

(define (sqrta x)
(sqrt-iter 1.0 x))

(define (square x)
(* x x))

(sqrta 9)

在R5RS下内存被耗尽,而在Lazy Racket下正常输出.

原因是new-if在R5RS,也就是应用序下,传入的每一个分支都会被求值.

<三>

(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))

(new-if #t (display "good") (display "bad"))

 R5RS输出:goodbad

 Lazy Racket输出:good