Solve Exercise 5-9: day of year functions with pointers

This exercise was fun and taught me a bit about initializing arrays and
working with pointers.

1 files changed, 74 insertions, 0 deletions

diff --git a/ch5/5-09_yearday-pointers.c b/ch5/5-09_yearday-pointers.c
new file mode 100644
index 0000000..7a28f88
--- /dev/null
+++ b/ch5/5-09_yearday-pointers.c
@@ -0,0 +1,74 @@
+#include <stdio.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 5-9: Rewrite the routines day_of_year() and month_day() with
+ * pointers instead of indexing.
+ */
+
+static char daytab[] = {
+	0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
+};
+
+int day_of_year(int year, int month, int day);
+void month_day(int year, int yearday, int *pmonth, int *pday);
+
+int main() {
+	int i, m, d, doy;
+	int tvals[][4] = {
+		{ 2012, 2, 29, 60 },
+		{ 2013, 3, 1, 60 },
+		{ 1985, 10, 22, 295 }
+	};
+	m = d = 0;
+	doy = 0;
+
+	// It said nothing about avoiding indexing in main()!
+	for (i = 0; i < 3; i++) {
+		doy = day_of_year(tvals[i][0], tvals[i][1], tvals[i][2]);
+		month_day(tvals[i][0], doy, &m, &d);
+		if (doy > 0 && m != 0 && d != 0) {
+			printf("%4d-%02d-%02d is day %3d of the year. ", tvals[i][0], tvals[i][1], tvals[i][2], doy);
+			printf("Day %3d of %4d is month %2d and day %3d.\n", doy, tvals[i][0], m, d);
+		} else {
+			printf("Out of range\n");
+		}
+	}
+	return 0;
+}
+
+int day_of_year(int year, int month, int day) {
+	int i, leap;
+	leap = (year % 4 == 0 && year % 100 != 0) || year % 400 == 0;
+
+	if (month < 1 || month > 12) {
+		return -1;
+	}
+	// Slightly complicated pointer math is needed to account for leap years
+	if (day > *(daytab + month) + ((month == 2) ? leap : 0) || day < 1) {
+		return -1;
+	}
+	for (i = 1; i < month; i++) {
+		// here, too
+		day += *(daytab + i) + ((i ==2) ? leap : 0);
+	}
+	return day;
+}
+
+void month_day(int year, int yearday, int *pmonth, int *pday) {
+	int i, leap;
+	leap = (year % 4 == 0 && year % 100 != 0) || year % 400 == 0;
+
+	if (yearday > 365 + leap || yearday < 1) {
+		*pmonth = 0;
+		*pday = 0;
+		return;
+	}
+	// ...and here. :P
+	for (i = 1; yearday > *(daytab + i) + ((i == 2) ? leap : 0); i++) {
+		yearday -= *(daytab + i) + ((i == 2) ? leap : 0);
+	}
+	*pmonth = i;
+	*pday = yearday;
+}
+