In this article, you will learn how to write a C program to check leap year. But let’s first understand what is a leap year before writing a C program to check whether a given year is a leap year or not.
What is a Leap Year?
A leap year is a calendar year that contains an additional day. It has 366 days instead of 365, by extending February to 29 days rather than the common 28. A leap year comes in every 4 years.
For example – 2004, 2008, 2012, 2016, 2020, etc.
You can easily check whether a given year is a leap or not using the C program. A year is a leap year if the following conditions match:
- If the given year is a multiple of 400. That means divided by 400.
- If the given year is a multiple of 4 but not a multiple of 100. That means divided by 4 but not divided by 100.
You just need to check the above condition for the leap year.
Algorithm to find leap year:
Following is the pseudo-code to find a leap year.
Algorithm: IF year MODULER 400 IS 0 THEN leap year ELSE IF year MODULER 100 IS 0 THEN not_leap_year ELSE IF year MODULER 4 IS 0 THEN leap_year ELSE not_leap_year
Explanation:
- If the given year is divisible by 400 then the first “if” statement executes and marks it leap year.
- If the first given year is not divisible by 400 then, the next else-if condition is checked, and if the given year is divisible by 100, then mark it as not a leap year.
- If the again else-if statement is not executed that means the given year is not divisible by 100 then, the next else-if condition is checked, and if the given year is divisible by 4, then mark it as a leap year.
- If none of the above is true then the else part executes and marks it as not a leap year.
Flow Chart to check Leap Year:
The below chart is describing the above-discussed pseudo-code to find the leap year.
C Program to check leap year:
Below we are writing some example programs that are describing how we can check leap year using the C language. So let us see the code,
Method 1:- Using an if-else statement
This is an implementation of the above pseudo-code.
#include <stdio.h>
int main()
{
unsigned int year = 0;
printf("Enter a valid year:");
scanf("%u",&year);
if((year%400 == 0) || ((year%4 == 0) && (year%100!=0)))
{
printf("%u is a leap year\n", year);
}
else
{
printf("%u is not a leap year\n", year);
}
return 0;
}
Output-1:
Enter a valid year:2020 2020 is a leap year
Output-2:
Enter a valid year:2001 2001 is not a leap year
Code Analysis:
In the above c code example, first, we ask the users to enter the year.
printf("Enter a valid year:");
scanf("%u",&year);
Now check the primary condition for a leap year. If the given year satisfies the condition of a leap year then the given year is a leap year.
if((year%400 == 0) || ((year%4 == 0) && (year%100!=0)))
{
printf("%u is a leap year\n", year);
}
else
{
printf("%u is not a leap year\n", year);
}
For example 2020. The year 2020 is divisible by 4 but not by 100. So, it is a leap year. But, the year 2001 is neither divisible by 4 nor by 400. So, it is not a leap year.
Method 2:- Using the if-else-if ladder
Another approach to checking leap year using the if-else-if statement.
#include <stdio.h>
int main()
{
unsigned int year=0;
printf("Enter a valid year:");
scanf("%u",&year);
if(((year%4) == 0) && ((year%100)!=0))
{
printf("%u is a Leap Year.\n\n\n",year);
}
else if(year%400 == 0)
{
printf("%u is a Leap Year.\n\n\n",year);
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
return 0;
}
Method 3:- Using the nested if-else
The third approach uses the nested if-else statement.
#include <stdio.h>
int main()
{
unsigned int year=0;
printf("Enter a valid year:");
scanf("%u",&year);
if ((year%4) == 0)
{
if ((year%100) == 0)
{
if ((year%400) == 0)
{
printf("%u is a Leap Year.\n\n\n",year);
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
}
else
{
printf("%u is a Leap Year.\n\n\n",year);
}
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
return 0;
}
Method 4:- Using Conditional Operator (ternary operator)
Here we check the leap year using the conditional operator.
#include <stdio.h>
int main()
{
unsigned int year=0;
printf("Enter the year:");
scanf("%u",&year);
//check leap year using conditional operator
int isLeapYear = ((year%400) != 0);
isLeapYear = isLeapYear? (((year%4) == 0) && ((year%100) != 0)):1;
if(isLeapYear)
{
printf("%u is a Leap Year.\n\n\n",year);
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
return 0;
}
Method 5:- Using a function
In this method, we create a function to find a leap year. It makes the code more readable. I think it is the best approach to check the leap year. In function, you can use any described method to check leap year. I am describing some ways to create a function to check leap year but there is “N way” you can create as per your requirement.
First Way:-
#include <stdio.h>
int isLeapYear(unsigned int year)
{
int result = 0;
if ((year%4) == 0)
{
if ((year%100) == 0)
{
if ((year%400) == 0)
{
result = 1;
}
}
else
{
result = 1;
}
}
return result;
}
int main()
{
unsigned int year=0;
int leapYear = 0;
printf("Enter the year:");
scanf("%u",&year);
//check leap year
leapYear = isLeapYear(year);
if(leapYear)
{
printf("%u is a Leap Year.\n\n\n",year);
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
return 0;
}
Output:
Enter the year:2010 2010 is Not a Leap Year.
Code Analysis:
In the above c code example, first, we ask the users to enter the year.
printf("Enter the year:");
scanf("%u",&year);
Now check the leap year to call the function isLeapYear(). If the year is leap year function returns 1 otherwise returns 0.
leapYear = isLeapYear(year);
if(leapYear)
{
printf("%u is a Leap Year.\n\n\n",year);
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
Second Way:
#include <stdio.h>
//Return 1 if year is leap year
int isLeapYear(unsigned int year)
{
int ret = ((year%400) != 0);
ret = ret? (((year%4) == 0) && ((year%100) != 0)):1;
return ret;
}
int main()
{
unsigned int year=0;
int leapYear = 0;
//Ask to enter year
printf("Enter the year:");
scanf("%u",&year);
//check leap year
leapYear = isLeapYear(year);
if(leapYear)
{
printf("%u is a Leap Year.\n\n\n",year);
}
else
{
printf("%u is Not a Leap Year.\n\n\n",year);
}
return 0;
}
C Program to Find Leap Years Within a Given Range:
In the previous examples, you learn how to check year is a leap year or not. But now I am asking to find leap years between the given two years. That means you need to find the leap years to the given range of years.
Don’t worry I am not only asking the question I am also writing the solution but I suppose you should try first.
#include <stdio.h>
//function to find leap year
int isLeapYear(unsigned int year)
{
int result = 0;
if ((year%4) == 0)
{
if ((year%100) == 0)
{
if ((year%400) == 0)
{
result = 1;
}
}
else
{
result = 1;
}
}
return result;
}
int main()
{
int startYear, endYear;
int year = 0;
printf("Enter the Valid Beginning year of the range: ");
scanf("%d",&startYear);
printf("Enter the Valid Ending year of the range: ");
scanf("%d",&endYear);
/*
Now find the leap year between
the starting and ending year.
*/
printf("Leap Years between %d and %d are: \n", startYear, endYear);
for (year= startYear; year<= endYear; ++year)
{
const int leapYearFound = isLeapYear(year);
if (leapYearFound)
{
printf("%d \n", year);
}
}
return 0;
}
Output:
Enter the Valid Beginning year of the range: 1991 Enter the Valid Ending year of the range: 2022 Leap Years between 1991 and 2022 are: 1992 1996 2000 2004 2008 2012 2016 2020
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