# SICP - Solution: Exercise 1.6

## Sep 28, 2018 21:03 · 192 words · 1 minute read

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.

Solution

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)
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.