Leap Year Program in C

When you write the leap year program in C you will understand about converting mathematical requirement into code logic.

It is important first to understand what a leap year is, it is then only we will be able to write a C program on it.

Let’s take a look.

Understanding a Leap Year

In a year where we all think of having 365 days, the earth’s year is actually 365 days, 5 hours, 48 minutes, and 46 seconds long.

Which is 11 minutes 14 seconds less than 6 hours in a year.

Which means, in 4 years, we will have an extra 23 hours, 15 minutes and 04 seconds, not really a complete 24 hours but still we see it as an extra day in 4 years.

But the short of approximate ~11 minutes in a year will make a short of ~1440 minutes (which is a day) in approximately ~131 years.

This is the reason we skip a leap year every century (100 years) and so almost 3 leap year days are skipped every ~400 years.

To make it clear, let’s understand this way. When the year 1600 is a leap year, then 1700, 1800, 1900 will NOT be leap years and 2000 will again become a leap year.

Understanding the Logic for Leap year C Program

Now let’s go to build a pseudo code before we actually write the C program.

Actual Logic

We will have to build some logic to rule out which numbers are not leap years. Then whichever is left out, lets see which is a leap year.

Based on the above understanding of a Leap Year, this is how we will do the same in C Program:

1. A year or a number which is divisible by 4 is generally a leap year.
   • In addition, a number which is divisible by 4 but also divisible by 400 is a leap year.
   • Yet again, a number which is divisible by 4 and also divisible by 100, is NOT a leap year.

Which means, any number that is divisible by 4 may NOT be a leap year. We still need to check further.

Which also means, any number that is divisible by 400 is straight forward a leap year.

But any number that is divisible by 4 but also divisible by 100 may not be a leap year.

Writing the Leap Year Program in C

To be able to understand this problem you must already know the below things in C:

• C datatypes
• C Conditional statement (if else or switch case)
• logical or condition (represent by the symbol ‘||’ in C)
• modulo operator ‘%’ in C

/*
 * Leap year program written by Sibananda Sahu
 * published on https://cprogram.dev on 4th June 2024
 * Last updated on 4th June 2024
 */

#include <stdio.h>

int main () {
    int year = 0;
    int is_leap_year = 1;

    printf ("Enter a number to check if it is a Leap Year or not: \n");
    scanf ("%d", &year);

    /* If the number is divisible by 4, there is a chance that it is a leap year */
    if ( year % 4 == 0 ) {
        /* A number divisible by 4 and 400 is a leap year */
        if ( year % 400 == 0 ) {
            printf ("Year %d is a LEAP YEAR\n", year);
        }

        /* A number divisible by 4 and 100 is NOT a leap year */
        else if ( year % 100 == 0 ) {
            printf ("Year %d is a NOT a leap year\n", year);
        }

        /* A number not divisible by 400 nor 100, but we already know it is divisible by 4 is a leap year */
        else {
            printf ("Year %d is a LEAP YEAR\n", year);
        }
    } // %4 if ends here but we have an else case to it

    /* A number that is not divisible by 4 is NOT a leap year. */
    else {
        printf ("Year %d is a NOT a leap year\n", year);
    } // %4 if else ends here

    return 0;
}Code language: PHP (php)

The same program can also be written as below:

/*
 * Leap year program written by Sibananda Sahu
 * published on https://cprogram.dev on 4th June 2024
 * Last updated on 4th June 2024
 */

#include <stdio.h>

int main () {
    int year = 0;
    int is_leap_year = 1;

    printf ("Enter a number to check if it is a Leap Year or not: \n");
    scanf ("%d", &year);

    /* If the number is divisible by 400, is a leap year */
    if ( year % 400 == 0 ) {
        printf ("Year %d is a LEAP YEAR\n", year);
    }

    /* A number NOT divisible by 400 and divisible by 100 is NOT a leap year */
    else if ( year % 100 == 0 ) {
        printf ("Year %d is a NOT a leap year\n", year);
    }

    /* A number not divisible by 400 nor 100, but divisible by 4 is a leap year */
    else if ( year % 4 == 0 ) {
        printf ("Year %d is a LEAP YEAR\n", year);
    }

    /* A number that is not divisible by 4 is NOT a leap year. */
    else {
        printf ("Year %d is a NOT a leap year\n", year);
    } // %400 if else ends here

    return 0;
}Code language: PHP (php)

Either of the program has the exact same logic but in two different forms and both of them work.

Output

$ ./a.out 
Enter a number to check if it is a Leap Year or not: 
1900
Year 1900 is a NOT a leap year

$ ./a.out
Enter a number to check if it is a Leap Year or not: 
2024
Year 2024 is a LEAP YEAR

$ ./a.out
Enter a number to check if it is a Leap Year or not: 
2000
Year 2000 is a LEAP YEAR

$ ./a.out
Enter a number to check if it is a Leap Year or not: 
2023
Year 2023 is a NOT a leap year

Conclusion

After completing the Leap Year C Program you will understand nested if else or if else in a better way. In addition, you will also understand using the modulo ‘%’ operator.

Understanding the real conditions before writing a program is necessary. Once you understand the actual mathematics then you may need to find which condition to keep and which one to rule out while building the program.

I have tested this Leap Year Program in C with several numbers and it worked pretty well. If you find any corner case that does not work, please leave that case in the comment I will fix the issue.

FAQs