# SICP - Solution: Exercise 1.6

##### September 28, 2018

## 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 #

Since `new-if`

is a function, and not a special form, each parameter subexpression will be evaluated *before* the procedure is applied. It means that when executing the `sqrt-iter`

function, `(good-enough? guess x)`

, `guess`

and `(sqrt-iter (improve guess x) x)`

will always be evaluated before `new-if`

is applied, whatever `(good-enough? guess x)`

evaluates to.

Since the second alternative `(sqrt-iter (improve guess x) x)`

is calling the function recursively, the function will be stuck in an infinite loop. In this case, it is interesting to note that `new-if`

will never be executed.

You can check by executing:

```
(define (new-if predicate then-clause else-clause)
(cond (predicate then-clause)
(else else-clause)))
(define (average x y) (/ (+ x y) 2))
(define (square x) (* x x))
(define (improve guess x) (average guess (/ x guess)))
(define (good-enough? guess x)
(< (abs (- (square guess) x)) 0.001))
(define (sqrt-iter guess x)
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x)))
(sqrt-iter 1.0 4) ; Infinite loop
```