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.
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