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#include <stdio.h>
/* The C Programming Language: 2nd Edition
*
* Exercise 5-8: There is no error checking in day_of_year or month_day.
* Remedy this defect.
*
*/
static char daytab[2][13] = {
{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{0, 31, 29, 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 m, d, doy;
m = d = 0;
doy = 0;
printf("Calling day_of_year(1973, 10, 32)\n> ");
if ((doy = day_of_year(1973, 10, 32)) > 0) {
printf("Oct 12nd, 1973 is day %d\n", doy);
} else {
printf("The month or day is out of range!\n");
}
printf("Calling month_day(1985, 295, ...)\n> ");
month_day(1985, 295, &m, &d);
if (m > 0 && d > 0) {
printf("Day 295 of year 1985 is month %d and day %d\n", m, d);
} else {
printf("month_day: day of year out of range!\n");
}
printf("Calling day_of_year(1492, 5, 14)\n> ");
if ((doy = day_of_year(1492, 5, 14)) > 0) {
printf("May 14th, 1492 is day %d\n", doy);
} else {
printf("The month or day is out of range!\n");
}
printf("Calling month_day(2012, 367, ...)\n> ");
month_day(2012, 367, &m, &d);
if (m > 0 && d > 0) {
printf("Day 295 of year 1985 is month %d and day %d\n", m, d);
} else {
printf("month_day: day of year 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;
}
if (day > daytab[leap][month] || day < 1) {
return -1;
}
for (i = 1; i < month; i++) {
day += daytab[leap][i];
}
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;
// Check for edge case
if (yearday > 365 + leap || yearday < 1) {
*pmonth = 0; // Set to zero so the error is obvious
*pday = 0;
return;
}
for (i = 1; yearday > daytab[leap][i]; i++) {
yearday -= daytab[leap][i];
}
*pmonth = i;
*pday = yearday;
}
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