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.
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Comments
It took me a while to get this new-if working in clojure, but here’s what I came up with. Any suggestions?
(def new-if (fn [predicate then-clause else-clause]
(cond predicate then-clause
:else else-clause)))
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