1.6

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)

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

Delighted, Alyssa uses new-if to rewrite the square-root program:

(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 (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 (new-sqrt x)
    (sqrt-iter 1.0 x)
)

(new-sqrt 9)

Timeout!

Because Scheme Lisp is in applicative order, so it evaluates all the parameters for the new-if; that makes sqrt-iter execute in whether cases, which cause infinite loop in consequences.

But the special form if will make sure the recursive happens only the predicate is not met.