C Program to Check Leap Year

Posted in C Tutorial

Updated on Oct 06, 2024

By Mari Selvan

👁️ 171 - Views

⏳ 4 mins

💬 1 Comment

Photo Credit to CodeToFun

🙋 Introduction

In the realm of programming, determining whether a given year is a leap year is a common requirement. A leap year is a year that is evenly divisible by 4, except for end-of-century years, which must be divisible by 400 to be a leap year. Checking for leap years involves implementing specific rules to validate the year.

In this tutorial, we will explore a C program designed to check whether a given year is a leap year.

📄 Example

Let's delve into the C code that accomplishes this task.

isLeapYear.c

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#include <stdio.h>

// Function to check if a year is a leap year
int isLeapYear(int year) {
  // Leap year conditions
  if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
    return 1; // True, it's a leap year
  else
    return 0; // False, it's not a leap year
}

// Driver program
int main() {
  // Replace this value with the year you want to check
  int year = 2024;

  // Call the function to check if the year is a leap year
  if (isLeapYear(year))
    printf("%d is a leap year.\n", year);
  else
    printf("%d is not a leap year.\n", year);

  return 0;
}

💻 Testing the Program

To test the program with different years, modify the value of year in the main program.

Output

2024 is a leap year.

🧠 How the Program Works

The program defines a function isLeapYear that takes an integer year as input and returns 1 (true) if the year is a leap year, and 0 (false) otherwise.
The function checks leap year conditions: the year must be divisible by 4 and not divisible by 100, or it must be divisible by 400.
Inside the main program, replace the value of year with the desired year you want to check.
The program calls the isLeapYear function and prints the result.

📏 Between the Given Range

Let's dive into the C code that checks for leap years in the specified range.

isLeapYear.c

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#include <stdio.h>

// Function to check if a year is a leap year
int isLeapYear(int year) {
  if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
    return 1; // Leap year
  } else {
    return 0; // Not a leap year
  }
}

// Driver program
int main() {
  // Define the range from 2000 to 2030
  int startYear = 2024;
  int endYear = 2050;

  printf("Leap years in the range %d to %d:\n", startYear, endYear);

  for (int year = startYear; year <= endYear; ++year) {
    if (isLeapYear(year)) {
      printf("%d ", year);
    }
  }

  return 0;
}

💻 Testing the Program

The provided code is already set to check for leap years in the range from 2024 to 2050.

Output

Leap years in the range 2024 to 2050:
2024 2028 2032 2036 2040 2044 2048

Compile and run the program to see the leap years within this range.

🧠 How the Program Works

The program defines a function isLeapYear that checks if a given year is a leap year using the rules mentioned earlier.
The main function initializes the range from 2000 to 2030 and iterates through each year.
For each year, it calls the isLeapYear function and prints the year if it is a leap year.

🧐 Understanding the Concept of Leap Year

Before delving into the code, let's understand the concept of leap years. Leap years are years that are evenly divisible by 4, except for end-of-century years, which must be divisible by 400 to be a leap year.