# SICP - Solution: Exercise 1.6

## Sep 28, 2018 21:03 · 192 words · 1 minute read

**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 parameters subexpressions will be evaluated *before* the procedure is applied. It means that when evaluating:

```
(new-if (good-enough? guess x)
guess
(sqrt-iter (improve guess x) x))
```

the predicate and the two alternatives will always be evaluated, whatever if `(good-enough? guess x)`

is evaluated to. Since the second alternative is calling the function itself recursively, the function will be stuck in an infinite loop.