C program to check whether a year is a leap year or not. Code: /* ** C program to check whether an entered year is a leap year or not ** @author: Pradeep ** @date: 02/12/06 */ #include<stdio.h> int main(void) { int year; printf("Enter the year: "); scanf("%d",&year); /* ** The logic is that the year is either divisible by both ** 100 and 4 , OR its only divisible by 4 not by hundred */ if(year%400 ==0 || (year%100 != 0 && year%4 == 0)) { printf("Year %d is a leap year",year); } else { printf("Year %d is not a leap year",year); } return 0; }

i want to know one thing that if yr is divisible by 400 than it will definately be divisible by 100 and if 1896 is leap yr than 1900 will definately be leap yr no checking reqd so from ma pt. of view hecking divisibilty by 4 is enough

No, it's not. Leap year does not always comes after 4 years. And yes, I did not do any typing mistake.

As per the current Gregorian calendar the determination of the leap year is as follows : All non-century years divisible by four are leap years. All century years divisible by 400 is a leap year. Which means 1900 & 2100 are not a leap years while year 2000 is a leap year. Read more here..

Functions to check for leap year in various languages: Code: function isLeapyear(year) { return year%400 ==0 || (year%100 != 0 && year%4 == 0); } PHP: function is_lear_year($year) { return $year%400 ==0 || ($year%100 != 0 && $year%4 == 0); } Code: sub{ $year = shift; return 1 if($year%400 ==0 || ($year%100 != 0 && $year%4 == 0)); # else ;-) return 0; } Code: Function IsLeapYear(y) If y Mod 400 = 0 Or (y Mod 100 <> 0 And y Mod 4 = 0) Then IsLeapYear = True Else IsLeapYear = False End If End Function

EXCELLENT post. I hadn't thought to check for leap this way. Saved me a lot of time with this math! Thanks!

Below is some code I wrote w/ pradeep's help. It was an assignment in a book I'm going through. --------------------------------------------------------------------------------------------- Code: #include<stdio.h> main() { int month, year, n, leap, notLeap; printf("\n\nEnter a month and year: "); scanf("%d %d", &month, &year); if(year % 400 == 0 || (year % 100 != 0 && year % 4 == 0)) year = leap; else year = notLeap; if (year == leap && month == 2) n = 29; if (year == notLeap && month == 2) n = 28; if (month == 1) n = 31; if (month == 3) n = 31; if (month == 4) n = 30; if (month == 5) n = 31; if (month == 6) n = 30; if (month == 7) n = 31; if (month == 8) n = 31; if (month == 9) n = 30; if (month == 10) n = 31; if (month == 11) n = 30; if (month == 12) n = 31; printf("\n\nThere are %d days in that month.", n); getch(); return; }

Your code is good but your logic isn't. Specifically, it's not just that the number is divisible by 100 and 4, it's that the number is divisible by 400. 200 is divisible by 100 and 4, but isn't a leap year. To say that a number is divisible by 100 and 4, or 100 or 4, is that same as saying a number is divisible by 4. That is: (A ^ B) v (A ^ !B) = A ^ (B v !B) = A. This is the same logic that makes all those ads so hilarious: "You can make up to $50,000 a year, or more!" So I guess the only amount you can't make is $50,000! Sorry, anal logician here -Plast