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.

The interpreter uses applicative-order in evaluating expressions. Since new-if is a procedure, all arguments must be evaluated before new-if itself can be evaluated. This means both then-clause and else-clause are run.

sqrt-iter is a recursive procedure, i.e. a procedure that calls itself. If the guess if good enough, it returns it, else it calls itself with an improved guess. However, in the case of Alyssa's implementation, sqrt-iter never returns because the recursion is in the else-clause in new-if. Applicative-order causes the else-clause be evaluated, thus making sqrt-iter to call itself again and again within new-if.