隨筆分類(157)

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)作為參數來傳遞給new-if，在執行new-if的時候，就總是會執行sqrt-iter (improve guess x)，造成了死循環。