Check Leap Year C Program

By | September 18, 2015

C Program To Check Leap Year

Learn How To Check Leap Year in C Programming Language. We have listed a well-researched Algorithm and Logic below for this program. To understand this program code efficiently, you must know How A Modulus Operator Works in C Programming Language. This code finds if a year entered by the User is a Leap Year or not using If Else Block Structure.

What is a Leap Year?

A Regular Year consists of 365 days. However, a Leap one consists of 366 days. It occurs after every three years and every fourth consecutive year is Leap. It consists of 52.2857 Weeks. This Leap Year C Program is based on the Gregorian Calendar.

Example

2004, 2008, 2012, 2016

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C Program To Check Leap Year using If Else

C Program To Find Leap Years from 1 To N

 

Output

Code To Check Leap Year in C Programming Language

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In case you get any Compilation Errors with this Code To Check Leap Year in C Programming Language or you have any doubt about it, let us know about it in the Comment Section below.

6 thoughts on “Check Leap Year C Program

  1. David Augusta

    Finally I got this code! This is just so easy to understand with all the conditions in different If Else blocks. Thank you so much!

    Reply
  2. Mahesh Kartik

    I was wondering if it is possible to find the date of the leap year in this C Program?

    Reply
    1. Tushar Soni Post author

      Yes. You can definitely find the leap year date in this C Code. However, that might make the code more complex, but its not that difficult. You need to find in a Library function that checks the date. There are some library functions available under the header .

      Reply
  3. Vijay Lumar

    This the most self explanatory code for leap year in C programming. I find it really easy to understand. Thanka for clearing my concepts about leap year.

    Reply
  4. Rajesh Srivastav

    The If-Else structure has helped a lot to understand the logic behind a C program to find if the year is a leap year or not. Thanks.

    Reply

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