C Program to find day of week given date

Question

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

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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     amp tm  0  sizeof tm      if  strptime str    d- m- Y    amp tm     NULL        time t t   mktime  amp tm       if  t  gt   0          return localtime  amp t - gt 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

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Not in one line of code  there s nothing for dealing with dates in the C standard library  It would be fairly simple to write a function based on the Doomsday algorithm  or similar  though

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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  lt  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  gt  gt  2     7        end rtc DayOfWeek

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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  amp  3     0   amp  amp    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  gt  gt   2     result    year   4     result -  year    25     result    year   100     result    year  gt  gt   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  lt   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

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Here s a C99 version based on wikipedia s article about Julian Day   include  lt stdio h gt   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

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include lt stdio h gt  int main void    int n y  int ly 0  int mon  printf  enter the date n    scanf   d   amp n   printf  enter the month in integer n    scanf   d   amp mon   mon mon-1  printf  enter year n    scanf   d   amp y   int dayT  dayT n 7  if  y 4  0 amp  amp y 100  0   y 4  0 amp  amp y 100  0 amp  amp 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 lt  3000 i       i i 99      if i lt 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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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  lt stdio h gt   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     amp d   amp m   amp y           correction for leap year     if  y   4    0  amp  amp   y   100    0    y   400    0           month 1    29       if  y  lt  1900    m  lt  1    m  gt  12    d  lt  1    d  gt  month m - 1             printf  This is an invalid date  n            return 1             for  i   1900  i  lt  y  i            if  i   4    0  amp  amp   i   100    0    i   400    0               d    366          else             d    365       for  i   0  i  lt  m - 1  i             d    month i        printf  This is a  s  n   day d   7        return 0

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Here is a simple code that I created in c that should fix your problem     include  lt conio h gt   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   amp y  amp month  amp 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  gt  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  gt  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

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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  lt  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     -

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include lt stdio h gt   include lt math h gt   include lt conio h gt  int fm int date  int month  int year    int fmonth  leap  if   year   100    0   amp  amp   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    amp year       printf   nEnter the month        scanf   d    amp month       printf   nEnter the date        scanf   d    amp 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

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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

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include lt stdio h gt  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   amp year  amp month  amp 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  amp  amp  year 100    0    year 400    0          if month  lt  1    month  gt 12          return -1          if  day  lt 1    day  gt  day tab leap  month           return -1          for i  1 i lt month   i                              day    day tab leap  year                     return day

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This one works  I took January 2006 as a reference   It is a Sunday   int isLeapYear int year          if   year 4  0  amp  amp  year 100  0      year 400  0             return 1       else          return 0       int isDateValid int dd int mm int yyyy         int isValid -1       if mm lt 0  mm gt 12            isValid -1            else        if  mm  1    mm  3    mm  5    mm  7    mm  8    mm  10    mm  12                if  dd gt 0  amp  amp  dd lt  31                isValid 1         else if  mm  4    mm  6    mm  9    mm  11              if  dd gt 0  amp  amp  dd lt  30                isValid 1         else                if isLeapYear yyyy                     if  dd gt 0  amp  amp dd lt 30                       isValid 1                 else                      if  dd gt 0  amp  amp dd lt 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 gt  2006 i--                             days   365 isLeapYear i                                                 printf  days after years is  d n  days                       for i mm-1 i gt 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

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Program to calculate the day on a given date by User      include lt stdio h gt   include lt conio h gt   include lt process h gt  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   amp dd  amp mm  amp 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 gt 29 amp  amp mm  2                             printf  You have entered wrong date                getch                exit 0                               else if dd gt 28 amp  amp 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 gt 29 amp mm  2                    printf  You have entered wrong date            getch            exit 0                    else if dd gt 28 amp  amp 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 gt 12    dd gt 31     yy gt 5000             printf  Your date is wrong        getch        exit 0            odd months should not contain more than 31 days if  dd gt 31  amp  amp   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 gt 30  amp  amp   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  amp  amp  remyr gt 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 lt mm-1 i                      odd1 odd1 montharr i                 todd odd1 dd      if todd gt 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

[c] C Program to find day of week given date

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