From d8c9ddf97c92d9dd3146a478c04b96b996d6d5c4 Mon Sep 17 00:00:00 2001
From: zlg <zlg@zlg.space>
Date: Sat, 20 Apr 2013 07:33:45 -0500
Subject: Solve Exercise 3-5: itob

---
 ch3/3-05_itob.c | 85 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 85 insertions(+)
 create mode 100644 ch3/3-05_itob.c

diff --git a/ch3/3-05_itob.c b/ch3/3-05_itob.c
new file mode 100644
index 0000000..58066e5
--- /dev/null
+++ b/ch3/3-05_itob.c
@@ -0,0 +1,85 @@
+#include <stdio.h>
+#include <string.h>
+#include <limits.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 3-5: Write the function itob(n,s,b) that converts the integer n
+ * into a base b character representation in the string s. In particular,
+ * itob(n,s,16) formats n as a hexadecimal integer in s.
+ *
+ * Answer: Retool the itoa() function written for 3-4 to take an additional
+ * argument, then add a few more checks and account for ASCII's 7-character
+ * distance between '9' and 'A', which makes bases higher than 16 function
+ * without a large switch statement.
+ */
+
+void reverse(char s[]) {
+	int c, i, j;
+
+	for (i = 0, j = strlen(s)-1; i < j; i++, j--) {
+		c = s[i];
+		s[i] = s[j];
+		s[j] = c;
+	}
+}
+
+void itob(int n, char s[], unsigned b) {
+	int i, sign, min, rem; // Add 'min' for later use
+
+	if ((sign = n) < 0) {
+		/* Detect this and add one so it can properly be made positive */
+		if (n == INT_MIN) {
+			n += 1;
+			min = 1;
+		}
+		n = -n;
+	}
+	i = 0;
+	if (b > 1) {
+		do {
+			/* If it's the first iteration of the loop and we've established
+			 * that n == INT_MIN, we need to add one to the resulting string for it
+			 * to be displayed properly. */
+			rem = n % b;
+			if (rem > 9) {
+				rem += 7;
+			}
+			if (i == 0 && min == 1) {
+				s[i++] = rem + '1';
+			} else {
+				s[i++] = rem + '0';
+			}
+		} while ((n /= b) > 0);
+		/* This is annoying, but to make hex numbers right, you have to account
+		 * for the +1 added to the final number causing bit flipping */
+		if (min == 1) {
+			for (i = 0; s[i] != '\0'; i++) {
+				if (s[i] == '-') {
+					continue;
+				}
+				if (i != 0 && s[i - 1] == '0' + (b + 7)) {
+					s[i] += 1;
+					s[i - 1] = '0';
+				}
+			}
+		}
+	}
+	if (sign < 0) {
+		s[i++] = '-';
+	}
+	s[i] = '\0';
+	reverse(s);
+}
+
+int main() {
+	int tests[5] = {INT_MIN, INT_MAX, -300, 172, 38478235};
+	char st[101] = "";
+	int i;
+	
+	for (i = 0; i < 5; i++) {
+		itob(tests[i], st, 16);
+		printf("%12d in string form is %12s\n", tests[i], st);
+	}
+	return 0;
+}
-- 
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