Last day of a given month - c

I've got an issue which I can't find a solution. Or at least a "good" one.
I want to find the last day of the month given a month and a year in C.
For example :
last_day(10, 2017) > 31
last_day(02, 2017) > 28
last_day(02, 2016) > 29
last_day(01, 2017) > 31
last_day(12, 2010) > 31
last_day(X, Y) > X is the month, Y the year
Here is my idea: Get the day on the month X + 1, of year Y. Remove 1 day from this date.
I would like to know if there is a better solution than that, since that will a make "lot" of operation for a "simple" thing.
Thanks.
Edit : https://ideone.com/sIISO1
#include <stdio.h>
#include <time.h>
#include <string.h>
int main(void) {
struct tm tm;
char out[256];
memset(&tm, 0, sizeof(struct tm));
tm.tm_mon = 1;
tm.tm_mday = 0;
strftime(out, 256, "%d-%m-%Y", &tm);
printf("%s", out);
return 0;
}
I've tested by using struct tm, and day = 0, in order to get the previous day but did not work

Ask point out in the commentary, I've complexify the problem way to much.
I have been inspired by what #Agnishom Chattopadhyay said in comment, which is get the date from a lookup table.
But I did make a function which did that
#include <stdio.h>
int days_in_month(int month, int year) {
if ( year < 1582 ) return 0; /* Before this year the Gregorian Calendar was not define */
if ( month > 12 || month < 1 ) return 0;
if (month == 4 || month == 6 || month == 9 || month == 11) return 30;
else if (month == 2) return (((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) ? 29 : 28);
return 31;
}
int main() {
printf("%d\n", days_in_month(10, 2017));
printf("%d\n", days_in_month(2, 2000));
printf("%d\n", days_in_month(2, 1300)); // Does not work !
printf("%d\n", days_in_month(2, 2018));
printf("%d\n", days_in_month(2, 2016));
}
https://ideone.com/5OZ3pZ

Related

Using macro as a range in C. Nested Loops. Logic

I am trying to make this program which takes YYYY and MM from the user. I have defined macros outside the main function, which is kinda range for years to be taken as input. For months, i have declared JAN and DEC as 1 and 12 respectively. What I am trying to is,show an error when something(month and year) is out of range, and the loop should repeat untill the right input has been entered. I tried my best to do that, but i didnt get anything. here is my code.
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#define MIN_YEAR 2012
#define MAX_YEAR 2022
int main(void)
{
const int JAN = 1;
const int DEC = 12;
int year = 0;
int month = 0;
do
{
printf("Set the year and month(YYYY MM): ");
scanf(" %d %d", &year, &month);
if ((MIN_YEAR > year && year > MAX_YEAR))
{
printf("ERROR: The year must be between 2012 and 2022 inclusive");
}
if ((JAN >= month >= DEC))
{
printf("ERROR: Jan.(1) - Dec.(12)");
}
else
{
printf("*** Log date set! ***");
}
} while (MIN_YEAR <= year <= MAX_YEAR , JAN <= month <= DEC);
return 0;
}
This condition
if ((MIN_YEAR > year && year > MAX_YEAR))
is logically incorrect. It seems you mean
if ( year < MIN_YEAR || year > MAX_YEAR )
The condition in this if statement
if ((JAN >= month >= DEC))
will always evaluate to false.
It is equivalent to
if ( (JAN >= month ) >= DEC )
So the sub-expression (JAN >= month ) evaluates either to 0 or 1 that in any case is less than DEC.
It seems you mean
if ( month < JAN || month > DEC )
In this condition
while (MIN_YEAR <= year <= MAX_YEAR , JAN <= month <= DEC)
there is used the comma operator. Its value is the values of the second operand JAN <= month <= DEC. That is as the first operand has no side effect then in fact the first operand MIN_YEAR <= year <= MAX_YEAR is ignored and does not influence on the result.
It seems you mean
while ( year < MIN_YEAR || year > MAX_YEAR || month < JAN || month > DEC)
Also there is another logical error
if ( ... )
{
printf("ERROR: The year must be between 2012 and 2022 inclusive");
}
if ( ... )
{
printf("ERROR: Jan.(1) - Dec.(12)");
}
else
{
printf("*** Log date set! ***");
}
The else statements belongs to the second if statement. So even if the year was entered incorrectly but the month was specified correctly then the else statement gets the control.
You should write for example
if ( ... )
{
printf("ERROR: The year must be between 2012 and 2022 inclusive");
}
else if ( ... )
{
printf("ERROR: Jan.(1) - Dec.(12)");
}
else
{
printf("*** Log date set! ***");
}

Calculate the end date from start date and number of days in C

for the past few hours I've been trying to figure out how to write a programme in C to calculate the end date based on the start date and number of days. (I haven't found the forum for this exact problem, yet).
So let's say you input 27/1/2021 as the starting date and then 380 days. The programme should now calculate and show you the end date 11/2/2022.
I don't know how to move forward, the help would be appreciated.
#include <stdio.h>
#include <stdlib.h>
int day, month, year, numberDays;
int leapYear(int year) {
return (year % 4 == 0 && (year % 100 != 0 || year % 400 == 0));
}
int monthYear[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main(void) {
printf("Enter starting date: ");
scanf("%d %d %d", &day, &month, &year);
printf("Enter number of days: ");
scanf("%d", &numberDays);
leapYear(year);
int wholeYears, rest;
if (leapYear(year)) {
wholeYears = numberDays / 366;
rest = numberDays % 366;
}
else {
wholeYears = numberDays / 365;
rest = numberDays % 365;
}
int resultYears = year + wholeYears;
int midDays = day + rest;
int resultMonths;
return 0;
}
I can't move any further. I'd need help.
So let's say you input 27/1/2021 as the starting date and then 380 days. The programme should now calculate and show you the end date 11/2/2022.
An easy approach is to use mktime() to bring a date into its standard range.
#include <stdio.h>
#include <time.h>
int main(void) {
struct tm start = {.tm_year = 2021 - 1900, .tm_mon = 1 - 1, .tm_mday = 27,
.tm_hour = 12}; // Midday to avoid DST issues.
start.tm_mday += 380;
printf("%d/%d/%d\n", start.tm_mday, start.tm_mon + 1, start.tm_year + 1900);
time_t t = mktime(&start);
printf("%d/%d/%d %s\n", start.tm_mday, start.tm_mon + 1, start.tm_year + 1900,
t == -1 ? "failed" : "OK");
}
Output
407/1/2021
11/2/2022 OK
Otherwise with discrete code, one is effectively re-writing the year/month/day portion of mktime().
I don't know how to move forward, the help would be appreciated.
Add the 380 to d of y, m, d. Maybe add some value to m in other cases. Then perform basic range reductions. There are 12 months to every year. Month range [1...12]. There are 365*400 + 97 days every 400 years. Day of the month is at least 1.
Last step is to handle a day outside the month range.
A simply, though inefficient, approach: test if the data is more than the days in that month. If so subtract the days in that month and advance to the next month.
I left out some code details as OP's goal is some ideas to move forward.
#define DAYSPER400YEARS (365 * 400 + 97)
#define JANUARY 1
#define FEBRUARY 2
#define DECEMBER 12
#define MONTHPERYEAR 12
static int is_leap_year(long long year) {
// If not divisible by 4 ...
// If not divisible by 100 ...
// If not divisible by 400 ...
}
static int days_per_month(long long year, int month) {
static const signed char dpm[] = { 0, //
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
if (month != FEBRUARY) {
return ....;
}
return .... + is_leap_year(year);
}
/*
* Bring a date into it primary range.
* https://en.wikipedia.org/wiki/Proleptic_Gregorian_calendar
* The year, month, day may be any value INT_MIN ... INT_MAX.
* Return error flag.
*/
bool ymd_to_primary_range(int *year, int *month, int *day) {
long long y = *year;
y += *month / 12;
*month %= 12; // month now in -11 to 11 range
while (*month < JANUARY) {
*month += MONTHPERYEAR;
year--;
}
y += (*day / DAYSPER400YEARS) * 400;
*day %= DAYSPER400YEARS;
while (*day < 1) {
*day += DAYSPER400YEARS;
y -= 400;
}
int dpm;
while (*day > (dpm = days_per_month(y, *month))) {
*day -= dpm;
(*month)++;
if (*month > ...) {
*month -= ....;
y++;
}
}
if (y < INT_MIN) {
*year = INT_MIN;
return true;
}
if (y > INT_MAX) {
*year = INT_MAX;
return true;
}
*year = (int) y;
return false;
}
Smaple
int main(void) {
int y = 2021;
int m = 1;
int d = 27;
d += 380;
printf("Error: %d\n", ymd_to_primary_range(&y, &m, &d));
printf("Date (dmy): %2d/%02d/%04d\n", d, m, y);
}
Output
Error: 0
Date (dmy): 11/02/2022

If inside a for loop (day difference for common and leap year) not working

I'm new to C-programming and I'm trying to make a code to count the day difference between a date and a birthday.
int year, birth_year;
int year_common;
int year_leap;
for(int i = year; i <= birth_year; i--){
year = i;
if (year % 400 != 0 || (year % 100 == 0 && year % 400 != 0)){
year_common++;
}
if(year % 400 == 0 || (year % 100 != 0 && year % 4 == 0)){
year_leap++;
}
}
int date_pass = (year_common * 365) + (year_leap * 366);
I wanted the loop to decrement the 'i', add 1 into the year_common integer when it's a common year, and add 1 into the year_leap integer when it's a leap year until it is the same as the birth_year.
For now, I'm still trying to check out the years, but no matter how many year difference I made, they always give out 5856 days.
e.g. :
birth date : 05/11/2005
checked date : 05/11/ 2007
give out '5856 days'
And I don't know where that number comes from.
I tried initializing it with 0, but it gave out 0 days instead.
I tried this code:
int year_pass = year - birth_year;
int year_leap = 0;
for(int i = year; i <= birth_year; i--){
if(i % 400 == 0 || (i % 100 != 0 && i % 4 == 0)){
year_leap++;
}
}
int date_pass = (year_pass * 365) + (year_leap * 1);
And it missed 1 day for the leap year.
Is there something wrong with the loop?
My prof wants the code to be as standard as possible with loops and conditions.
You have declared the year common and year leap as in variables but haven't assigned any value. Since youre adding one unit by ++;, you need to pre define year_leap and year_common =0
int year, birth_year;
int year_common=0;
int year_leap=0;
Dates are hard to handle, with all UTC leap seconds and local timezones - leap years are only the tip of the iceberg. Don't reinvent the wheel. Whenever you want to do something with dates, first thing to do is to convert everything to seconds since epoch. Then you do what you want.
#include <time.h>
#include <stdio.h>
#include <stdint.h>
int main() {
// 05/11/2005
struct tm birthday = {
.tm_year = 2005 - 1900,
.tm_mon = 11,
.tm_mday = 5,
.tm_hour = 12,
};
// 05/11/2007
struct tm end = {
.tm_year = 2007 - 1900,
.tm_mon = 11,
.tm_mday = 5,
.tm_hour = 12,
};
time_t birth_s = mktime(&birthday);
time_t end_s = mktime(&end);
time_t diff_s = end_s - birth_s;
time_t diff_day = diff_s / 3600 / 24;
printf("%ju\n", (uintmax_t)diff_day);
}

How to check if date is valid using mktime in C?

I have a little problem with my code. I need to check if user's given date is valid or not. I want to check it using 2 rules:
1. Date is valid date and found in a calendar (for example 29.2.2015 is invalid because there's only 28 days in February 2015) 2. Day and month can only be max 2 numbers (for example day 10 and 02 are valid but 010 and 002 are not). All help is appreciated!
This is my code this far:
void dateValidator(const char *date1) {
struct tm date = {0};
int day1;
int month1;
int year1;
int vday;
int vmonth;
int vyear;
sscanf(date1, "%3d.%3d.%d",&day1,&month1,&year1);
/**How do I check that if sscanf reads more than 2 characters on month and day, date is
invalid?**/
date.tm_year = year1 - 1900;
date.tm_mon = month1 - 1;
date.tm_mday = day1;
date.tm_isdst = -1;
vday = date.tm_mday;
vmonth = date.tm_mon;
vyear = date.tm_year;
mktime(&pvm);
if ((vday != date.tm_mday) || (vmonth != date.tm_mon) || (vyear != date.tm_year)) {
/**This doesnt seem to work**/
printf("Invalid date");
}
}
I have deleted my previous answer and posting another one. Actually your code is fine, You just have to send the correct parameter to mktime and to check its return value. I.e. change:
mktime(&pvm);
if ((vday != date.tm_mday) || (vmonth != date.tm_mon) || (vyear != date.tm_year)) {
/**This doesnt seem to work**/
printf("Invalid date");
}
to something like:
r = mktime(&date);
if (r == -1 || (vday != date.tm_mday) || (vmonth != date.tm_mon) || (vyear != date.tm_year)) {
printf("Invalid date");
}
and declare the variable r as time_t r;.
int IsValidDate(int year, int month, int day)
{
unsigned int leap;
unsigned char mon_day[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
/* check the rang of the year */
if ((year < 1) || (year >= 3200))
{
return 0;
}
if ((month < 1) || (month > 12))
{
return 0;
}
/* if it's leep year */
if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))
{
leap = 1;
}
else
{
leap = 0;
}
mon_day[1] += leap;
if ((day > mon_day[month - 1]) || (day < 1))
{
return 0;
}
return 1;
}
For ruler 1, I have wrote this code for check the date before.
And I hope this code can help you.
The ruler 2 can easily checked, you can check it by format the valid date and compare it with the input string.
How do I check that if sscanf reads more than 2 characters on month and day, date is invalid?
Day and month can only be max 2 numbers digits (for example day 10 and 02 are valid but 010 and 002 are not).
An easy check: Record the offset at the end of the scan and tests 1) that it is not the original 0 (scanning got that far) and 2) this is the end of the string so as to detect extra junk in data1[]. Change 3d to 2d.
int n = 0;
sscanf(date1, "%2d.%2d.%d %n",&day1,&month1,&year1, &n);
if (n > 0 && data1[n] == '\0') Success();
else Fail();
Pedantic code would check with "%[]" to disallow leading spaces and '+'.
int n = 0;
sscanf(date1, "%*2[0-9].%*2[0-9].%*4[0-9] %n", &n);
if (n == 0 || data1[n]) Fail();
else {
sscanf(date1, "%2d.%2d.%d",&day1,&month1,&year1);
...
How to check if date is valid using mktime?
#Marian posted a good solution which adds a test of the return value of mktime().
if (-1 == mktime(&data)) Fail();
if (vday != date.tm_mday) || (vmonth != date.tm_mon) || (vyear != date.tm_year) Fail();

C Program to find day of week given date

Is there a way to find out day of the week given date in just one line of C code?
For example
Given 19-05-2011(dd-mm-yyyy) gives me Thursday
As reported also by Wikipedia, in 1990 Michael Keith and Tom Craver published an expression to minimise the number of keystrokes needed to enter a self-contained function for converting a Gregorian date into a numerical day of the week.
The expression does preserve neither y nor d, and returns a zero-based index representing the day, starting with Sunday, i.e. if the day is Monday the expression returns 1.
A code example which uses the expression follows:
int d = 15 ; //Day 1-31
int m = 5 ; //Month 1-12`
int y = 2013 ; //Year 2013`
int weekday = (d += m < 3 ? y-- : y - 2, 23*m/9 + d + 4 + y/4- y/100 + y/400)%7;
The expression uses the comma operator, as discussed in this answer.
Enjoy! ;-)
A one-liner is unlikely, but the strptime function can be used to parse your date format and the struct tm argument can be queried for its tm_wday member on systems that modify those fields automatically (e.g. some glibc implementations).
int get_weekday(char * str) {
struct tm tm;
memset((void *) &tm, 0, sizeof(tm));
if (strptime(str, "%d-%m-%Y", &tm) != NULL) {
time_t t = mktime(&tm);
if (t >= 0) {
return localtime(&t)->tm_wday; // Sunday=0, Monday=1, etc.
}
}
return -1;
}
Or you could encode these rules to do some arithmetic in a really long single line:
1 Jan 1900 was a Monday.
Thirty days has September, April, June and November; all the rest have thirty-one, saving February alone, which has twenty-eight, rain or shine, and on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
EDIT: note that this solution only works for dates after the UNIX epoch (1970-01-01T00:00:00Z).
Here's a C99 version based on wikipedia's article about Julian Day
#include <stdio.h>
const char *wd(int year, int month, int day) {
/* using C99 compound literals in a single line: notice the splicing */
return ((const char *[]) \
{"Monday", "Tuesday", "Wednesday", \
"Thursday", "Friday", "Saturday", "Sunday"})[ \
( \
day \
+ ((153 * (month + 12 * ((14 - month) / 12) - 3) + 2) / 5) \
+ (365 * (year + 4800 - ((14 - month) / 12))) \
+ ((year + 4800 - ((14 - month) / 12)) / 4) \
- ((year + 4800 - ((14 - month) / 12)) / 100) \
+ ((year + 4800 - ((14 - month) / 12)) / 400) \
- 32045 \
) % 7];
}
int main(void) {
printf("%d-%02d-%02d: %s\n", 2011, 5, 19, wd(2011, 5, 19));
printf("%d-%02d-%02d: %s\n", 2038, 1, 19, wd(2038, 1, 19));
return 0;
}
By removing the splicing and spaces from the return line in the wd() function, it can be compacted to a 286 character single line :)
This is my implementation. It's very short and includes error checking. If you want dates before 01-01-1900, you could easily change the anchor to the starting date of the Gregorian calendar.
#include <stdio.h>
int main(int argv, char** arv) {
int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
char* day[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};
int d, m, y, i;
printf("Fill in a date after 01-01-1900 as dd-mm-yyyy: ");
scanf("%d-%d-%d", &d, &m, &y);
// correction for leap year
if (y % 4 == 0 && (y % 100 != 0 || y % 400 == 0))
month[1] = 29;
if (y < 1900 || m < 1 || m > 12 || d < 1 || d > month[m - 1]) {
printf("This is an invalid date.\n");
return 1;
}
for (i = 1900; i < y; i++)
if (i % 4 == 0 && (i % 100 != 0 || i % 400 == 0))
d += 366;
else
d += 365;
for (i = 0; i < m - 1; i++)
d += month[i];
printf("This is a %s.\n", day[d % 7]);
return 0;
}
The answer I came up with:
const int16_t TM_MON_DAYS_ACCU[12] = {
0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334
};
int tm_is_leap_year(unsigned year) {
return ((year & 3) == 0) && ((year % 400 == 0) || (year % 100 != 0));
}
// The "Doomsday" the the day of the week of March 0th,
// i.e the last day of February.
// In common years January 3rd has the same day of the week,
// and on leap years it's January 4th.
int tm_doomsday(int year) {
int result;
result = TM_WDAY_TUE;
result += year; // I optimized the calculation a bit:
result += year >>= 2; // result += year / 4
result -= year /= 25; // result += year / 100
result += year >>= 2; // result += year / 400
return result;
}
void tm_get_wyday(int year, int mon, int mday, int *wday, int *yday) {
int is_leap_year = tm_is_leap_year(year);
// How many days passed since Jan 1st?
*yday = TM_MON_DAYS_ACCU[mon] + mday + (mon <= TM_MON_FEB ? 0 : is_leap_year) - 1;
// Which day of the week was Jan 1st of the given year?
int jan1 = tm_doomsday(year) - 2 - is_leap_year;
// Now just add these two values.
*wday = (jan1 + *yday) % 7;
}
with these defines (matching struct tm of time.h):
#define TM_WDAY_SUN 0
#define TM_WDAY_MON 1
#define TM_WDAY_TUE 2
#define TM_WDAY_WED 3
#define TM_WDAY_THU 4
#define TM_WDAY_FRI 5
#define TM_WDAY_SAT 6
#define TM_MON_JAN 0
#define TM_MON_FEB 1
#define TM_MON_MAR 2
#define TM_MON_APR 3
#define TM_MON_MAY 4
#define TM_MON_JUN 5
#define TM_MON_JUL 6
#define TM_MON_AUG 7
#define TM_MON_SEP 8
#define TM_MON_OCT 9
#define TM_MON_NOV 10
#define TM_MON_DEC 11
#include<stdio.h>
#include<math.h>
#include<conio.h>
int fm(int date, int month, int year) {
int fmonth, leap;
if ((year % 100 == 0) && (year % 400 != 0))
leap = 0;
else if (year % 4 == 0)
leap = 1;
else
leap = 0;
fmonth = 3 + (2 - leap) * ((month + 2) / (2 * month))+ (5 * month + month / 9) / 2;
fmonth = fmonth % 7;
return fmonth;
}
int day_of_week(int date, int month, int year) {
int dayOfWeek;
int YY = year % 100;
int century = year / 100;
printf("\nDate: %d/%d/%d \n", date, month, year);
dayOfWeek = 1.25 * YY + fm(date, month, year) + date - 2 * (century % 4);
//remainder on division by 7
dayOfWeek = dayOfWeek % 7;
switch (dayOfWeek) {
case 0:
printf("weekday = Saturday");
break;
case 1:
printf("weekday = Sunday");
break;
case 2:
printf("weekday = Monday");
break;
case 3:
printf("weekday = Tuesday");
break;
case 4:
printf("weekday = Wednesday");
break;
case 5:
printf("weekday = Thursday");
break;
case 6:
printf("weekday = Friday");
break;
default:
printf("Incorrect data");
}
return 0;
}
int main() {
int date, month, year;
printf("\nEnter the year ");
scanf("%d", &year);
printf("\nEnter the month ");
scanf("%d", &month);
printf("\nEnter the date ");
scanf("%d", &date);
day_of_week(date, month, year);
return 0;
}
OUTPUT:
Enter the year 2012
Enter the month 02
Enter the date 29
Date: 29/2/2012
weekday = Wednesday
This is question 20.31 in the C FAQ list. It lists three answers, two of which are amenable to one-liners.
Modified Zeller's congruence: j = y/100; k = y%100; dow = (d + 26 * (m + 1) / 10 + k + k/4 + j/4 + 5*j) % 7;
Tomohiko Sakamoto's method: static int t[] = {0,3,2,5,0,3,5,1,4,6,2,4}; y -= m < 3; dow = (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
There's a further condensation of Sakamoto's method using an ASCII string instead of the t array: dow = (y + y/4 - y/100 + y/400 + "-bed=pen+mad."[m] + d) % 7;
The basic idea underlying all of these methods is the same: boil the date down to a monotonically-increasing sequence of day numbers (taking month lengths and leap years into account), then reduce that day number modulo 7.
I think you can find that in glib:
http://developer.gnome.org/glib/unstable/glib-Date-and-Time-Functions.html#g-date-get-day
Regards
#include<stdio.h>
static char day_tab[2][13] = {
{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0,31,29,31,30,31,30,31,31,30,31,30,31}
};
int main()
{
int year,month;
scanf("%d%d%d",&year,&month,&day);
printf("%d\n",day_of_year(year,month,day));
return 0;
}
int day_of_year(int year ,int month,int day)
{
int i,leap;
leap = year%4 == 0 && year%100 != 0 || year%400 == 0;
if(month < 1 || month >12)
return -1;
if (day <1 || day > day_tab[leap][month])
return -1;
for(i= 1;i<month ; i++)
{
day += day_tab[leap][year];
}
return day;
}
/*
Program to calculate the day on a given date by User
*/
#include<stdio.h>
#include<conio.h>
#include<process.h>
void main()
{
int dd=0,mm=0,i=0,yy=0,odd1=0,todd=0;//variable declaration for inputing the date
int remyr=0,remyr1=0,lyrs=0,oyrs=0,cyr=0,upyr=0,leap=0;//variable declaration for calculation of odd days
int montharr[12]={31,28,31,30,31,30,31,31,30,31,30,31};//array of month days
clrscr();
printf("Enter the date as DD-MM-YY for which you want to know the day\t:");
scanf("%d%d%d",&dd,&mm,&yy); //input the date
/*
check out correct date or not?
*/
if(yy%100==0)
{
if(yy%400==0)
{
//its the leap year
leap=1;
if(dd>29&&mm==2)
{
printf("You have entered wrong date");
getch();
exit(0);
}
}
else if(dd>28&&mm==2)
{
//not the leap year
printf("You have entered wrong date");
getch();
exit(0);
}
}
else if(yy%4==0)
{
//again leap year
leap=1;
if(dd>29&mm==2)
{
printf("You have entered wrong date");
getch();
exit(0);
}
}
else if(dd>28&&mm==2)
{
//not the leap year
printf("You have entered wrong date");
getch();
exit(0);
}
//if the leap year feb month contains 29 days
if(leap==1)
{
montharr[1]=29;
}
//check date,month,year should not be beyond the limits
if((mm>12)||(dd>31)|| (yy>5000))
{
printf("Your date is wrong");
getch();
exit(0);
}
//odd months should not contain more than 31 days
if((dd>31 && (mm == 1||mm==3||mm==5||mm==7||mm==8||mm==10||mm==12)))
{
printf("Your date is wrong");
getch();
exit(0);
}
//even months should not contains more than 30 days
if((dd>30 && (mm == 4||mm==6||mm==9||mm==11)))
{
printf("Your date is wrong");
getch();
exit(0);
}
//logic to calculate odd days.....
printf("\nYou have entered date: %d-%d-%d ",dd,mm,yy);
remyr1=yy-1;
remyr=remyr1%400;
cyr=remyr/100;
if(remyr==0)
{
oyrs=0;
}
else if(cyr==0 && remyr>0)
{
oyrs=0;
}
else if(cyr==1)
{
oyrs=5;
}
else if(cyr==2)
{
oyrs=3;
}
else if(cyr==3)
{
oyrs=1;
}
upyr=remyr%100;
lyrs=upyr/4;
odd1=lyrs+upyr;
odd1=odd1%7;
odd1=odd1+oyrs;
for(i=0;i<mm-1;i++)
{
odd1=odd1+montharr[i];
}
todd=odd1+dd;
if(todd>7)
todd=todd%7; //total odd days gives the re quired day....
printf("\n\nThe day on %d-%d-%d :",dd,mm,yy);
if(todd==0)
printf("Sunday");
if(todd==1)
printf("Monday");
if(todd==2)
printf("Tuesday");
if(todd==3)
printf("Wednesday");
if(todd==4)
printf("Thrusday");
if(todd==5)
printf("Friday");
if(todd==6)
printf("Saturday");
getch();
}
For Day of Week, years 2000 - 2099.
uint8_t rtc_DayOfWeek(uint8_t year, uint8_t month, uint8_t day)
{
//static const uint8_t month_offset_table[] = {0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}; // Typical table.
// Added 1 to Jan, Feb. Subtracted 1 from each instead of adding 6 in calc below.
static const uint8_t month_offset_table[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
// Year is 0 - 99, representing years 2000 - 2099
// Month starts at 0.
// Day starts at 1.
// Subtract 1 in calc for Jan, Feb, only in leap years.
// Subtracting 1 from year has the effect of subtracting 2 in leap years, subtracting 1 otherwise.
// Adding 1 for Jan, Feb in Month Table so calc ends up subtracting 1 for Jan, Feb, only in leap years.
// All of this complication to avoid the check if it is a leap year.
if (month < 2) {
year--;
}
// Century constant is 6. Subtract 1 from Month Table, so difference is 7.
// Sunday (0), Monday (1) ...
return (day + month_offset_table[month] + year + (year >> 2)) % 7;
} /* end rtc_DayOfWeek() */
This one works: I took January 2006 as a reference. (It is a Sunday)
int isLeapYear(int year) {
if(((year%4==0)&&(year%100!=0))||((year%400==0)))
return 1;
else
return 0;
}
int isDateValid(int dd,int mm,int yyyy) {
int isValid=-1;
if(mm<0||mm>12) {
isValid=-1;
}
else {
if((mm==1)||(mm==3)||(mm==5)||(mm==7)||(mm==8)||(mm==10)||(mm==12)) {
if((dd>0)&&(dd<=31))
isValid=1;
} else if((mm==4)||(mm==6)||(mm==9)||(mm==11)) {
if((dd>0)&&(dd<=30))
isValid=1;
} else {
if(isLeapYear(yyyy)){
if((dd>0)&&dd<30)
isValid=1;
} else {
if((dd>0)&&dd<29)
isValid=1;
}
}
}
return isValid;
}
int calculateDayOfWeek(int dd,int mm,int yyyy) {
if(isDateValid(dd,mm,yyyy)==-1) {
return -1;
}
int days=0;
int i;
for(i=yyyy-1;i>=2006;i--) {
days+=(365+isLeapYear(i));
}
printf("days after years is %d\n",days);
for(i=mm-1;i>0;i--) {
if((i==1)||(i==3)||(i==5)||(i==7)||(i==8)||(i==10)) {
days+=31;
}
else if((i==4)||(i==6)||(i==9)||(i==11)) {
days+=30;
} else {
days+= (28+isLeapYear(i));
}
}
printf("days after months is %d\n",days);
days+=dd;
printf("days after days is %d\n",days);
return ((days-1)%7);
}
Here is a simple code that I created in c that should fix your problem :
#include <conio.h>
int main()
{
int y,n,oy,ly,td,a,month,mon_,d,days,down,up; // oy==ordinary year, td=total days, d=date
printf("Enter the year,month,date: ");
scanf("%d%d%d",&y,&month,&d);
n= y-1; //here we subtracted one year because we have to find on a particular day of that year, so we will not count whole year.
oy= n%4;
if(oy==0) // for leap year
{
mon_= month-1;
down= mon_/2; //down means months containing 30 days.
up= mon_-down; // up means months containing 31 days.
if(mon_>=2)
{
days=(up*31)+((down-1)*30)+29+d; // here in down case one month will be of feb so we subtracted 1 and after that seperately
td= (oy*365)+(ly*366)+days; // added 29 days as it is the if block of leap year case.
}
if(mon_==1)
{
days=(up*31)+d;
td= (oy*365)+(ly*366)+days;
}
if(mon_==0)
{
days= d;
td= (oy*365)+(ly*366)+days;
}
}
else
{
mon_= month-1;
down= mon_/2;
up= mon_-down;
if(mon_>=2)
{
days=(up*31)+((down-1)*30)+28+d;
td= (oy*365)+(ly*366)+days;
}
if(mon_==1)
{
days=(up*31)+d;
td= (oy*365)+(ly*366)+days;
}
if(mon_==0)
{
days= d;
td= (oy*365)+(ly*366)+days;
}
}
ly= n/4;
a= td%7;
if(a==0)
printf("\nSunday");
if(a==1)
printf("\nMonday");
if(a==2)
printf("\nTuesday");
if(a==3)
printf("\nWednesday");
if(a==4)
printf("\nThursday");
if(a==5)
printf("\nFriday");
if(a==6)
printf("\nSaturday");
return 0;
}
I wrote a small function to do this in C inspired by Conway's doomsday algorithm.
It was written with 8-bit microcontrollers in mind and the "year" argument is a year from 0 to 99 only (representing 2000 to 2099). I wrote it to compile down nice and small on avr-gcc. It uses no lookup table (I got the code smaller without).
// day of week, Thomas Rutter
// y should be a year 0 to 99 (meaning 2000 to 2099)
// m and d are 1-based (1-12, 1-31)
// returns 0 = sunday, ..., 6 = saturday
uint8_t dayofweek(uint8_t y, uint8_t m, uint8_t d)
{
uint8_t x = y + 24 - (m < 3);
x = x + (x >> 2) + d - m;
if (m & 1)
x += (m & 8) ? 4 : 3;
if (m < 3)
x += 3;
return x % 7;
}
In my own version, the mod operator at the bottom was replaced with my own optimised code as the platform I was using has no MUL or DIV and I wanted it 1 to 7 instead of 0 to 6.
This could be adapted to work with other centuries or specify the year in full - I just didn't need to do so for my own use.
See strftime and %u or %w qualifiers
#include<stdio.h>
int main(void) {
int n,y;
int ly=0;
int mon;
printf("enter the date\n");
scanf("%d",&n);
printf("enter the month in integer\n");
scanf("%d",&mon);
mon=mon-1;
printf("enter year\n");
scanf("%d",&y);
int dayT;
dayT=n%7;
if((y%4==0&&y%100!=0)|(y%4==0&&y%100==0&&y%400==0))
{
ly=y;
printf("the given year is a leap year\n");
}
char a[12]={6,2,2,5,0,3,5,1,4,6,2,4};
if(ly!=0)
{
a[0]=5;
a[1]=1;
}
int m,p;
m=a[mon];
int i,j=0,t=1;
for(i=1600;i<=3000;i++)
{
i=i+99;
if(i<y)
{
if(t==1)
{
p=5;t++;
}
else if(t==2)
{
p=3;
t++;
}
else if(t==3)
{
p=1;
t++;
}
else
{
p=0;
t=1;
}
}}
int q,r,s;
q=y%100;
r=q%7;
s=q/4;
int yTerm;
yTerm=p+r+s;
int w=dayT+m+yTerm;
w=w%7;
if(w==0)
printf("SUNDAY");
else if(w==1)
printf("MONDAY");
else if(w==2)
printf("TUESDAY");
else if(w==3)
printf("WEDNESDAY");
else if(w==4)
printf("THURSDAY");
else if(w==5)
printf("FRIDAY");
else
printf("SATURDAY");
return 0;
}

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