SICP Exercise 1.6

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

The code is here:


Explanation: new-if is a procedure. So when it is called, it tries to evaluate all its arguments first (applicative order evaluation). One of the arguments is 'new-sqrt-iter' so this results in recursive calls that never end. 'new-if' does not fully execute and return even once.

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