Solve Exercise 3-6: itoa (3 arg version)

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diff --git a/ch3/3-06_itoa3.c b/ch3/3-06_itoa3.c
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+++ b/ch3/3-06_itoa3.c
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+#include <stdio.h>
+#include <string.h>
+#include <limits.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 3-6: Write a version of itoa that accepts three arguments instead
+ * of two. The third argument is a minimum field width; the converted number
+ * must be padded with blanks on the left if necessary to make it wide enough.
+ *
+ * Answer: This is trivial. Process the number, then check the value of i. If
+ * it's less than the field width, add a blank until i is equal to the field
+ * width and terminate your string.
+ */
+
+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[], unsigned fw) {
+	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++] = '-';
+	}
+	while (i < fw) {
+		s[i++] = ' ';
+	}
+	s[i] = '\0';
+	reverse(s);
+}
+
+int main() {
+	int i;
+	char foo[16] = "";
+	int tests[5] = {-25, 409689, 8, -1000, 135};
+	for (i = 0; i < 5; i++) {
+		itoa(tests[i], foo, 5);
+		printf("Pad to 5 spaces! '%s'\n", foo);
+	}
+	return 0;
+}