Re: [Haskell-beginners] find element of tupels

thx Daniel. Now I implemented it like that to use my monthAndMaxDay list and just use the isLeapYear check for February: legalDate :: Date -> Bool legalDate (myDay, Feb, myYear) = 0 < myDay && myDay <= (if isLeapYear myYear then 29 else 28) legalDate (myDay, myMonth, myYear) = maybe False id $ do days <- lookup myMonth monthAndMaxDay return (not (myDay <= 0) && (myDay <= days)) -------- Original-Nachricht --------

Datum: Fri, 18 Dec 2009 00:17:56 +0100 Von: Daniel Fischer An: beginners@haskell.org Betreff: Re: [Haskell-beginners] find element of tupels

Am Donnerstag 17 Dezember 2009 20:44:34 schrieb Henk-Jan van Tuyl:

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On Thu, 17 Dec 2009 17:18:41 +0100, wrote:

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thanks Magnus, great, works so far. I added to check that myDay have to be greater 0:

legalDate :: Date -> Bool legalDate (myDay, myMonth, myYear) = maybe False id $ do days <- lookup myMonth monthAndMaxDay return (not (myDay <= 0) && (myDay <= days))

Another thing is, that I have to check for leapYears. I have a function that checks if a year is a leap year

isLeapYear :: Int -> Bool isLeapYear year = mod year 4 == 0

That is not entirely correct; from package time: isLeapYear :: Integer -> Bool isLeapYear year = (mod year 4 == 0) && ((mod year 400 == 0) || not (mod year 100 == 0))

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monthAndMaxDay has 30 days for February which have to be fix. So i have to check in legalDate if the year is a leap year. If so I have to check myDay in February for 29 days, otherwise for 28 days. How can I implement this to isLegalDate?

Use fromGregorianValid from package time.

Or

legalDate (d,Feb,y) = 0 < d && d < (if isLeapYear y then 30 else 29) legalDate (d,m,y) | m `elem` [Apr,Jun,Sep,Nov] = 0 < d && d < 31 | otherwise = 0 < d && d < 32

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Regards, Henk-Jan van Tuyl

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