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Leap year 20213/20/2023 ![]() This means that centuries are only a leap year if they are divisible by 400. Note: The above rule does not apply to century years.Ĭenturies such as 19 only have a leap day if they are divisible by 400.ġ900 is divisible by 4 and also by 100, but not by 400, so not a leap year. By adding once in the 4 years one extra additional day this problem is solved.Īny year that is divisible by 4 is a leap year, such as 2016, 2020, 2024, 2028. This is done because one year doesn't contain 365 days but 365.25 days. When is the next leap year?A leap year is a year with 366 days instead of 365, every 4 years in February one extra day is added. datecheck.In the table below you can see all leap years between 20. Sample session: $ gnatmake -gnatwa datecheck.adb &. Put_Line(IsValidateDate(Dates(i))'Image) The first rule most of us know is that a leap year occurs every 4 years, which adds an extra day to the February calendar. Dates(4) := (2000, FEB, 32) - raises warning, fails with constraint error The test for valid day then is thus reduced to a single function call Date.Day 31,īegin return (if Month = FEB then GetFebMaxDay(Year) else MaxDaysInMonth(Month)) end įunction IsValidateDate(Date: DateT) return Boolean is ![]() 6 8 02:56 20 years old level / An office worker / A public employee / Very /. writing story, needed to check if future year was a leap year on the fly. Return (if IsLeapYear(Year) then 29 else 28) Its cool to see the algo logic that makes it all work :) 5 1 15:18 30 years old level / - / Very /. I have used the new Ada equivalent to the C ternary operator cond?a:b, (if cond then a else b): function GetFebMaxDay(Year: YearT) return DayInMonthT is That treatment has again been stowed away in its own little subroutine GetFebMaxDay(Day, Month), making everything neat and intelligible. In the Gregorian calendar, three conditions are used to identify leap years: The year can be evenly divided by 4, is a leap year, unless: The year can be evenly divided by 100, it is NOT a leap year, unless: The year is also evenly divisible by 400. Only days in February must be treated specially. ![]() Pitfall 3: Trying to use the Pay Period Leap Year for other. In the end, they got the math right when the employee pointed it out and they made the employee whole. ![]() This is the pitfall the reader’s company fell into. The year is necessary only in the case of February, where the maximum allowed day of the month depends on whether it is in a leap year.įor all other months the return value is constant, which I have recorded in a const array MaxDaysInMonth. 2021 (a Pay Period Leap Year): 13 payments of 1925.93 (last day is June 30, first payroll includes 2 weeks of 2020) corrected. In the code below I have a routine GetMaxDayInMonth(Month, Year) that gives the maximum day for the given month in the given year. I think a lot can be gained by putting certain tests in subroutines, even if they are only called once. My concern is the if statement where I put Day = 29 and Month = 2, can this be solved without having the need to put two if statements? If Year mod 400 = 0 or (Year mod 4 = 0 and Year mod 100 /= 0) then Month = 8 or Month = 10 or Month = 12) then This is my code (only the part I need help with): Get(Days) Įlsif Day = 31 and (Month = 2 or Month = 4 or Month = 6 or Month = 9 or Month = 11) thenĮlsif Day = 30 and (Month = 1 or Month = 3 or Month = 5 or Month = 7 or If it's a leap year, February has 29 days instead of 28. Years are also evenly dividable by 400 leap years. Last years wildfires burned a portion of the Stags Leap District winerys property. Information about months and leap years: Month: Jan Feb Mar Apr May Jun Jul Aug Sep Okt Nov Decĭays : 31 28/29 31 30 31 30 31 31 30 31 30 30Ī year is a leap year if it's evenly dividable by 4, but not even when divided by 100. Descargar esta imagen: A trio of goats graze on a hilltop at Ilsley Vineyards to help create fire breaks in Napa, Calif., on Tuesday, March 23, 2021. My program runs but I don't like how my last if statement looks. I'm not going to post the entire code because it'll be too much. I'm having a problem with handling the leap year in a good way. I'm supposed to do a program that handles the dates of a year.
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