SICP - Solution: Exercise 1.6

SICP - Solution: Exercise 1.6

September 28, 2018

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
  (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)
          (sqrt-iter (improve guess x) x)))

What happens when Alyssa attempts to use this to compute square roots? Explain.


Since new-if is a function, and not a special form, each parameters subexpressions will be evaluated before the procedure is applied. It means that when evaluating:

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

the predicate and the two alternatives will always be evaluated, whatever if (good-enough? guess x) is evaluated to. Since the second alternative is calling the function itself recursively, the function will be stuck in an infinite loop.