Exercise 1.6.  Alyssa P. Hacker doesn't see why if needs to be provided as a special form. ``Why
can't I just define it as an ordinary procedure in terms of cond?'' she asks. Alyssa's friend Eva Lu
Ator claims this can indeed be done, and she defines a new version of if:
(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))
Eva demonstrates the program for Alyssa:
(new-if (= 2 3) 0 5)
5
(new-if (= 1 1) 0 5)
0
Delighted, Alyssa uses new-if to rewrite the square-root program:
32(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x)
                     x)))
What happens when Alyssa attempts to use this to compute square roots? Explain.


(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
      (else-clause)))
(define (average x y)
  (/ (+ x y) 2))
(define (improve guess x)
  (average guess (/ x guess)))
(define (good-enough? guess x)
  (< (abs (- (square guess) x))0.001))
(define (square x)
  (* x x))
(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x)
                     x)))

sqrt-iter (improve guess x)作为参数来传递给new-if,在执行new-if的时候,就总是会执行sqrt-iter (improve guess x),造成了死循环。

posted on 2009-03-06 15:19 lzj520 阅读(296) 评论(0)  编辑  收藏 所属分类: 个人学习日记sicp

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