Re: [Haskell-cafe] how can I do this the best

From the homework:

data MessageType = Info | Warning | Error Int deriving (Show, Eq) data LogMessage = LogMessage MessageType TimeStamp String deriving (Eq, Show) data MaybeLogMessage = ValidM LogMessage -- A valid msg | InvalidLM String -- Invalid msg parseMessage :: String -> MaybeLogMessage parseMessage = undefined To implement parseMessage, we consume the string from left to right word-by-word. If the first word is E, then we read the second word as in integer indicating severity and proceed further. Info and Warning don't require more information, so the next word will be the timestamp in those cases. If the pattern fails anywhere, we return the whole string as an InvalidLM. Hope this helps. On 24 February 2015 at 06:37, Richard A. O'Keefe wrote:

On 24/02/2015, at 5:19 am, Roelof Wobben wrote:

...

I tried it another way more like explained on this page : http://www.seas.upenn.edu/~cis194/spring13/lectures/02-ADTs.html

so I tried this :

parseMessage :: [Char] -> [Char] parseMessage s case Errornumber of IsDigit Errornumber -> "Geldige string" otherwise -> "Ongeldige string" where Error = s words Errornumber = Error(ErrorNumber _ _ ) Errorcode = Error(_ Errorcode _ )

but now I cannot use where :(

That's not your problem.

IsDigit ErrorNumber is not a pattern.

parseMessage s = if isDigit errorNumber then "Geldige string" else "Ongelidige string" where errorNumber = ???

is OK.

Now I cannot make sense of Error = s words

identifiers beginning with capital letters are used for - module names - type constructors - data constructors You want a variable here, so it must begin with a lower case letter.

s words treats a string s as a function and applies it to the function words as argument: s(words). But that does not type check. You mean words s.

The result of words s, whatever else it may be, is not an error.

Errornumber = Error(ErrorNumber _ _)

In the form "expr where pattern = expr", the thing after the equal sign must be an expression. But Error(ErrorNumber _ _) is not an expression. "_" is a PATTERN (= I do not care what goes here) but never an EXPRESSION (because what value would it have?).

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-- Regards Sumit Sahrawat