(sqrt-iter 1.0 x))
;; 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?''
+(defn new-if [predicate then-clause else-clause]
+ (cond predicate then-clause
+ :else else-clause))
+
+(new-if (= 3 2) 0 5) ; 5
+(new-if (= 1 1) 0 5) ; 0
+
+;; Delighted, Alyssa uses new-if to rewrite the square-root program:
+
+(defn 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.
+(comment
+ Since `new-if' is a function, when it is called from sqrt-iter, the parameters
+ are evaluated before it gets called. good-enough? will return a false unless
+ the guess and x are almost the same. guess evaluated to the initial value of
+ guess. sqrt-iter gets evaluated, but gets into an infinite loop. The predicate
+ will have no effect.)
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