How to find leap year programatically in C

Question

I made a program using C to find whether the entered year is a leap year or not. But unfortunately its not working well. It says a year is leap and the preceding year is not leap.

#include<stdio.h>
#include<conio.h>
int yearr(int year);
void main(void)
{
    int year;
    printf("Enter a year:");
    scanf("%d",&year);
    if(!yearr(year))
    {
        printf("It is a leap year.");
    }
    else
    {
    printf("It is not a leap year");
    }


getch();
}
int yearr(int year)
{
    if((year%4==0)&&(year/4!=0))
    return 1;
    else
    return 0;
}

After reading the comments i edited my coding as:

#include<stdio.h>
#include<conio.h>
int yearr(int year);
void main(void)
{
    int year;
    printf("Enter a year:");
    scanf("%d",&year);
    if(!yearr(year))
    {
        printf("It is a leap year.");
    }
    else
    {
    printf("It is not a leap year");
    }


getch();
}
int yearr(int year)
{
    if((year%4==0)
    {
    if(year%400==0)
    return 1;
    if(year%100==0)
    return 0; 
    }
    else
    return 0;
}

Answer 1

I used this code:

#include <stdio.h>

int main()
{
    int yr;
    printf ("Enter a year \n");
    scanf ("%d", &yr);

    if (yr%400 == 0)
        printf("\n LEAP YEAR.");

    else if (yr%4==0 && yr%100!=0)
        printf("\n LEAP YEAR.");
    else
        printf ("\n NOT LEAP YEAR.");
}

Answer 2

#define is_leap(A) !((A) & 3)

Just make sure you don't enter negative year :)

Answer 3

This could be the right solution. Algorithm given on Wikipedia is not right.

-(BOOL)isLeapYear: (int)year{    

    if(year%4==0){
      if(year%100!=0){
        return YES;
     }
     else if(year%400!=0){
        return YES;
     }
     else return NO;
   }

    else return NO;
  }

Answer 4

Kevin's answer provides an optimal 8 operation test (with XOR using constants) but if you are looking for something a bit more readable, try this 9 operation test.

year % 4 == 0 && !((year % 100 == 0) ^ (year % 400 == 0))

Truth table for (year % 100 == 0) ^ (year % 400 == 0)

                              (year % 100 == 0) ^ (year % 400 == 0)
100 doesnt divide year     .    F
only 100 divides year      .    T
100 and 400 divides year   .    F

Now !(year % 100 == 0) ^ (year % 400 == 0) gives what you want.

Answer 5

http://www.wwu.edu/depts/skywise/leapyear.html

Leap Year Rules

There is a leap year every year whose number is perfectly divisible by four - except for years which are both divisible by 100 and not divisible by 400. The second part of the rule effects century years. For example; the century years 1600 and 2000 are leap years, but the century years 1700, 1800, and 1900 are not. This means that three times out of every four hundred years there are eight years between leap years.

Answer 6

    #include 
    void main(void)
    {
        int year;
        printf("Enter a year to check if it is Leap Year\n");
        scanf("%d",&year);
        if(year%400==0) /* Why  mod 400 */
            printf("%d is a Leap Year\n",year);
        else if(year%100==0) /*  Why  mod 100  */
            printf("%d is not a Leap Year\n",year);
        else if(year%4==0)
            printf("%d is a Leap Year\n",year);
        else
            printf("%d is not a Leap Year\n",year);

    }