In this code snippet, we’ll show an example of a C language function to find (Tomorrows date) the next date in the Gregorian Calendar.

A C language function to find (Tomorrows date) the next date in the Gregorian Calendar.

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#include <stdio.h>
#include <stdlib.h>
int isLeapYear(int year);
void findNextDate(int day, int month, int year);
int isLeapYear(int year) {
if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
return 1; // The year is a leap year
} else {
return 0; // The year is not a leap year
}
}
void findNextDate(int day, int month, int year) {
int daysInMonth[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// Check if the year is a leap year and adjust the number of days in February
if (month == 2 && isLeapYear(year)) {
daysInMonth[month] = 29;
}
// Check the validity of the date
if (month < 1 || month > 12 || day < 1 || day > daysInMonth[month]) {
printf("The entered date is not valid.\n");
return;
}
if (day < daysInMonth[month]) {
day++;
} else {
day = 1;
month++;
if (month > 12) {
month = 1;
year++;
}
}
printf("The next date is: %02d/%02d/%d\n", day, month, year);
}
int main() {
int day, month, year;
printf("Enter the date (day month year): ");
scanf("%d %d %d", &day, &month, &year);
findNextDate(day, month, year);
return 0;
}
#include <stdio.h> #include <stdlib.h> int isLeapYear(int year); void findNextDate(int day, int month, int year); int isLeapYear(int year) { if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) { return 1; // The year is a leap year } else { return 0; // The year is not a leap year } } void findNextDate(int day, int month, int year) { int daysInMonth[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; // Check if the year is a leap year and adjust the number of days in February if (month == 2 && isLeapYear(year)) { daysInMonth[month] = 29; } // Check the validity of the date if (month < 1 || month > 12 || day < 1 || day > daysInMonth[month]) { printf("The entered date is not valid.\n"); return; } if (day < daysInMonth[month]) { day++; } else { day = 1; month++; if (month > 12) { month = 1; year++; } } printf("The next date is: %02d/%02d/%d\n", day, month, year); } int main() { int day, month, year; printf("Enter the date (day month year): "); scanf("%d %d %d", &day, &month, &year); findNextDate(day, month, year); return 0; }
#include <stdio.h>
#include <stdlib.h>

int isLeapYear(int year);

void findNextDate(int day, int month, int year);

int isLeapYear(int year) {
    if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
        return 1; // The year is a leap year
    } else {
        return 0; // The year is not a leap year
    }
}

void findNextDate(int day, int month, int year) {
    int daysInMonth[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    // Check if the year is a leap year and adjust the number of days in February
    if (month == 2 && isLeapYear(year)) {
        daysInMonth[month] = 29;
    }

    // Check the validity of the date
    if (month < 1 || month > 12 || day < 1 || day > daysInMonth[month]) {
        printf("The entered date is not valid.\n");
        return;
    }

    if (day < daysInMonth[month]) {
        day++;
    } else {
        day = 1;
        month++;
        if (month > 12) {
            month = 1;
            year++;
        }
    }
    printf("The next date is: %02d/%02d/%d\n", day, month, year);
}

int main() {
    int day, month, year;

    printf("Enter the date (day month year): ");
    scanf("%d %d %d", &day, &month, &year);

    findNextDate(day, month, year);

    return 0;
}

MIT Licensed Code – See License

Tags: C, tomorrows date, date calculations, timedate, date, time, c method, stdio, stdlib