# 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)
```

```
(new-if (= 1 1) 0 5)
```

## 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.

```
(define (improve guess x)
(average guess (/ x guess)))
(define (average x y)
(/ (+ x y) 2))
(define (good-enough? guess x)
(< (abs (- (square guess) x)) 0.001))
```

```
(define (new-sqrt x)
(sqrt-iter 1.0 x)
)
```

```
(new-sqrt 9)
```

Timeout!

Because Scheme Lisp is in applicative order, so it evaluates all the parameters for the `new-if`

; that makes `sqrt-iter`

execute in whether cases, which cause infinite loop in consequences.

But the special form `if`

will make sure the recursive happens only the predicate is not met.