sub abs_day { # this subroutine accepts a month, day, and two-digit year in # mm/dd/yy numerical form, e.g., 3/23/04. # it then calculates the "absolute day", # i.e., the number of days # since day 0, which is the turn of the 21st century. # so, 1/10/2000 is absolute day 10; # 1/10/2001 is absolute day 376 # (because year 2000 was a leap year). # the routine deals with leap years by # multiplying years by 365.25, and then rounding down. my $month = shift(@_); my $day = shift(@_); my $year = shift(@_); if ($debug) { print "i'm in abs_day and month is $month,", " day is $day, year is $year\n"; } my $abs_day; undef($abs_day); # so each year has 365.25 days # years are of the form 00, 01, 02, etc. $abs_day = $year * 365.25; # add .75 to account for fact that year 00 is a leap year $abs_day = $abs_day + .75; if ($debug) { print "abs_day is $abs_day after adding .75 to yr*365\n"; } if ($month == 1) #jan { $abs_day = $abs_day + 0; } if ($month == 2) #feb { $abs_day = $abs_day + 31; } # starting in march, we add .25 to number of days # to account for leap year if ($month == 3) #mar { $abs_day = $abs_day + 59.25; } if ($month == 4) #apr { $abs_day = $abs_day + 90.25; } if ($month == 5) #may { $abs_day = $abs_day + 120.25; } if ($month == 6) #jun { $abs_day = $abs_day + 151.25; } if ($month == 7) #jul { $abs_day = $abs_day + 181.25; } if ($month == 8) #aug { $abs_day = $abs_day + 212.25; } if ($month == 9) #sep { } if ($month == 11) #nov { $abs_day = $abs_day + 304.25; } if ($month == 12) #dec { $abs_day = $abs_day + 334.25; } $abs_day = $abs_day + $day; if ($debug) { print "now that we have added year, month, and", " day, abs_day is $abs_day\n"; } $abs_day = sprintf "%d", $abs_day; if ($debug) { print "now that we have converted from float to integer,", " abs_day is $abs_day\n"; } return $abs_day; }