From: Ramakrishnan Muthukrishnan <vu3rdd@gmail.com>
Date: Thu, 25 Mar 2010 02:41:56 +0000 (+0530)
Subject: Solution to exercise 1.6.
X-Git-Url: https://git.rkrishnan.org/uri/frontends/FTP-and-SFTP.rst?a=commitdiff_plain;h=6a7606d4a71f8eed73e5883311d60524512e709c;p=sicp.git

Solution to exercise 1.6.
---

diff --git a/chapter1/ch1_1.clj b/chapter1/ch1_1.clj
index 733f23c..777e384 100644
--- a/chapter1/ch1_1.clj
+++ b/chapter1/ch1_1.clj
@@ -115,3 +115,27 @@
   (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.)
\ No newline at end of file