ex-1.6.md

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.

if and cond are special forms because of their own evaluation rules. Nto all operands given to if and cond are evaluated.

While new-if is an ordinary procedure. All operands given to new-if are evaluated at first. And the body of sqrt-iter recursively calls itself. Therefore, the sqrt-iter defined with new-if will never end.