Solve Exercise 4-12: Recursive itoa()

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diff --git a/ch4/4-12_recursive-itoa.c b/ch4/4-12_recursive-itoa.c
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+#include <stdio.h>
+#include <string.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 4-12: Adapt the ideas of printd() to write a recursive version of
+ * itoa; that is, convert an integer into a string by calling a recursive
+ * routine.
+ *
+ * Answer: Making itoa() recursive isn't too much work; it helps knowing how to
+ * use static variables; they make it possible to increment the position in the
+ * string without using an external variable. Additionally, you'll need to
+ * understand flow control to prevent the sub-routines from executing further
+ * than needed. As a result, itoa() gets called multiple times, but if it
+ * calls itself, it immediately returns so it won't venture further. This is
+ * to make sure i is reset to zero only after the number is finished, so the
+ * next call starts at the beginning of the string.
+ *
+ * Normally, recursive functions shouldn't worry about state, but it's necessary
+ * in this version of itoa.
+ */
+
+void itoa(int, char[]);
+void reverse(char[]);
+
+int main(void) {
+	char foo[40] = "";
+
+	itoa(829048, foo);
+	printf("%s\n", foo);
+	itoa(-4021, foo);
+	printf("%s\n", foo);
+
+	return 0;
+}
+
+void itoa(int num, char target[]) {
+	static int i = 0;
+	static int neg = 0;
+
+	if (num < 0) {
+		neg = 1;
+		num = -num;
+	}
+
+	if (num /= 10 > 0) {
+		target[i++] = (num % 10) + '0';
+		num /= 10;
+		itoa(num, target);
+		return;
+	} else {
+		if (neg == 1) {
+			neg = 0;
+			target[i++] = '-';
+			target[i] = '\0';
+		} else {
+			target[i] = '\0';
+		}
+		reverse(target);
+	}
+	i = 0;
+}
+
+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;
+	}
+}