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C++ Program to Check Leap Year

Posted in C++ Tutorial
Updated on Oct 06, 2024
By Mari Selvan
πŸ‘οΈ 99 - Views
⏳ 4 mins
πŸ’¬ 1 Comment
C++ Program to Check Leap Year

Photo Credit to CodeToFun

πŸ™‹ 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.

πŸ“„ Example

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

isLeapYear.cpp
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#include <iostream>

// Function to check if a year is a leap year
bool isLeapYear(int year) {
  // Leap year conditions
  if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
    return true; // True, it's a leap year
  else
    return false; // 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))
    std::cout << year << " is a leap year." << std::endl;
  else
    std::cout << year << " is not a leap year." << std::endl;

  return 0;
}

πŸ’» 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.

Compile and run the program to check if the specified year is a leap year.

🧠 How the Program Works

  1. The program defines a function isLeapYear that takes an integer year as input and returns true if the year is a leap year, and false otherwise.
  2. 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.
  3. Inside the main program, replace the value of year with the desired year you want to check.
  4. The program calls the isLeapYear function and prints the result using std::cout.

πŸ“ Between the Given Range

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

isLeapYear.cpp
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#include <iostream>

// Function to check if a year is a leap year
bool isLeapYear(int year) {
  return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
}

// Driver program
int main() {
  std::cout << "Leap Years in the range 2024 to 2050:" << std::endl;

  // Iterate over the range and check for leap years
  for (int year = 2024; year <= 2050; ++year) {
    if (isLeapYear(year)) {
      std::cout << year << " ";
    }
  }

  std::cout << std::endl;

  return 0;
}

πŸ’» 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 in this range.

🧠 How the Program Works

  1. The program defines a function isLeapYear that takes a year as input and returns true if it is a leap year, and false otherwise.
  2. The main function iterates through the range from 2024 to 2050 and prints the years that meet the leap year criteria.

🧐 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.

🎒 Optimizing the Program

While the provided program is effective, consider exploring and implementing alternative approaches or optimizations for checking leap years.

Feel free to incorporate and modify this code as needed for your specific use case. Happy coding!

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Author

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πŸ‘‹ Hey, I'm Mari Selvan

For over eight years, I worked as a full-stack web developer. Now, I have chosen my profession as a full-time blogger at codetofun.com.

Buy me a coffee to make codetofun.com free for everyone.

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Mari Selvan
Mari Selvan
10 months ago

If you have any doubts regarding this article (C++ Program to Check Leap Year), please comment here. I will help you immediately.

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