Solve Exercise 2-3: Hex to integer converter

This exercise was fun, and I learned a simpler way to convert them.

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diff --git a/ch2/2-03_hex-to-int.c b/ch2/2-03_hex-to-int.c
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+#include <stdio.h>
+#include <ctype.h>
+
+/* The C Programming Language, 2nd Edition
+ *
+ * Exercise 2-3: Write the function htoi(s), which converts a string of
+ * hexadecimal digits (including an optional 0x or 0X) into its equivalent
+ * integer value. The allowable digits are 0 through 9, a through f, and A
+ * through F.
+ *
+ * Answer: ctype.h has some cool functions in it. Using just two functions from
+ * that header, my function's length is much shorter than it would be if I
+ * were to wing it alone with C primitives.
+ *
+ * Anyway, the idea is to make sure you're working with a hex digit,
+ * homogenize the case of the letters, and ignore x's. Since hex is base-16
+ * instead of base-10, plug a 16 in and perform some clever math based on
+ * ASCII knowledge.
+ *
+ * Since the function can be objectively proven correct, I included test cases
+ * to ensure that its output is correct.
+ *
+ * `man ascii` for more info, if you're using an OS that's actually useful.
+ */
+
+int htoi(char s[]) {
+	int i, val;
+
+	for (i = val = 0; isxdigit(s[i]) || toupper(s[i]) == 'X'; ++i) {
+		if (toupper(s[i]) == 'X') {
+			continue;
+		}
+		if (s[i] > '9') {
+			val = 16 * val + (toupper(s[i]) - '7');
+			// The 7 is because 'A' is 7 higher than '9' in ASCII and thus only needs
+			// to be knocked down by that much to fall in line with the normal integer
+			// conversion
+		} else {
+			val = 16 * val + (s[i] - '0');
+		}
+	}
+	return val;
+}
+
+int main() {
+	printf(" HEX | DECIMAL\n---------------\n   1 = %3d\n 0xf = %3d\n0X64 = %3d\n  fF = %3d\n 093 = %3d\n", htoi("1"), htoi("0xf"), htoi("0X64"), htoi("fF"), htoi("093"));
+	return 0;
+}