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 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:
(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.
new-if is a function, and not a special form, each parameter subexpression will be evaluated before the procedure is applied. It means that when evaluating:
(new-if (good-enough? guess x) guess (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.