SICP 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 problem with the sqrt-iter is that new-if will evaluate forever. Functions use applicative-order evaluation, whereas conditionals use normal order (or something more similar to normal order). When new-if is called it will attempt to evaluate a version of sqrt-iter, which itself calls new-if, causing infinite evaluations.