Exercise 1.6

文責:@naoiwata
(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)

(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 (square x) (* x x))

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

(sqrt 3)
new-if は通常の手続きなので各要素を評価していく.
よって sqrt-iter における new-if の評価で無限再帰呼び出しとなる.

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Exercise 1.7