Solve Exercise 3-4: itoa improved

1 files changed, 71 insertions, 0 deletions

diff --git a/ch3/3-04_itoa2.c b/ch3/3-04_itoa2.c
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+++ b/ch3/3-04_itoa2.c
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+#include <stdio.h>
+#include <string.h>
+#include <limits.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 3-4: In a two's complement number representation, our version of
+ * itoa does not handle the largest negative number, that is, the value of n
+ * equal to -(2^wordsize-1). Explain why not. Modify it to print that value
+ * correctly, regardless of the machine on which it runs.
+ *
+ * Answer: The reason it doesn't work is because zero (0) takes up one of the
+ * possible numbers. When the number n is INT_MIN, the operation `n = -n` to
+ * make it positive doesn't work, which makes the math later on moot. I fixed
+ * this by detecting that it's the lowest possible number, then adding one.
+ * This makes the resulting positive number the largest possible positive
+ * number. In order to make the string appear right, I spend the first
+ * iteration of the do/while loop detecting INT_MIN and adding '1' to the
+ * character that results from the math. The string then outputs properly.
+ */
+
+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 itoa(int n, char s[]) {
+	int i, sign, min; // 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;
+	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. */
+		if (i == 0 && min == 1) {
+			s[i++] = n % 10 + '1';
+		} else {
+			s[i++] = n % 10 + '0';
+		}
+	} while ((n /= 10) > 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++) {
+		itoa(tests[i], st);
+		printf("%12d in string form is %12s\n", tests[i], st);
+	}
+	return 0;
+}