As Alex mentioned, isValid returns Bool, while type for putStrLn is `String -> IO ()`. So, in order to print something of type Bool, you need to first convert it to String. For example, via a function `show`:

putStrLn (show True)

As Alex mentioned, there's a `print` function, which does exactly this:

print x = putStrLn (show x)

You can use it.

On Mon, Feb 23, 2015 at 7:19 PM, Roelof Wobben <r.wobben@home.nl> wrote:

And when Im trying this:

{-# OPTIONS_GHC -Wall #-}

module LogAnalysis where

import Log;
import Data.Char (isLetter, isDigit)

isValid :: String -> Bool
isValid s = go (words s)
    where
      go ([a]:b:_) = isLetter a && all isDigit b
      go _         = False

   
-- | The main entry point.
main :: IO ()
main = do
    putStrLn $ isValid "I 4764 He trusts to you to set them free,"


I see this error message :

src/LogAnalysis.hs@19:16-19:67

Couldn't match type

Bool

with

[Char]

Expected type: String Actual type: Bool … In the second argument of ‘($)’, namely ‘isValid "I 4764 He trusts to you to set them free,"’ In a stmt of a 'do' block: putStrLn $ isValid "I 4764 He trusts to you to set them free,"

Roelof

Roelof Wobben schreef op 23-2-2015 om 17:19:

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 :(

Roelof




Roelof Wobben schreef op 23-2-2015 om 16:10:

Oke,

Then I make there a mistake,

What I try to do is to send the file to parseMessage and let IsValid check if it´s have the right format.

Then after the where I try to check if the function isValid returns true or false.

Roelof


Konstantine Rybnikov schreef op 23-2-2015 om 16:03:

Roelof,

You defined isValid function in the upper-scope first, and then you defined a symbol (variable) that re-wrote that name to something different (string "Geldige string"). That's why you get an error saying it doesn't expect arguments.

My suggestion is to rename second isValid.

Good luck.

On Mon, Feb 23, 2015 at 4:50 PM, Roelof Wobben <r.wobben@home.nl> wrote:

Chaddaï Fouché schreef op 23-2-2015 om 13:20:

Note that Roelof is doing the CIS 194 Homework http://www.seas.upenn.edu/~cis194/fall14/hw/03-ADTs.pdf (the older version of fall2014, not the one currently running). This is much clearer than Roelof's description, and gives among other information an algebraic datatype to represent log messages.


--

Jedaï

Correct and Im trying to do exercise 1 of Week 2,

I have tried this solution :

-- | Main entry point to the application.
{-# OPTIONS_GHC -Wall #-}

module LogAnalysis where

import Log;
import Data.Char (isLetter, isDigit)

isValid :: [Char] -> Bool
isValid s = go (words s)
    where
      go ([a]:b:_) = isLetter a && all isDigit b
      go _         = False

parseMessage :: [Char] -> [Char]
parseMessage s = isValid s
    where
        isValid = "Geldige string"
        _       = "Ongeldige string"

-- | The main entry point.
main :: IO ()
main = do
    putStrLn $ parseMessage "I 4764 He trusts to you to set them free,"


but I see this error message :

src/LogAnalysis.hs@16:18-16:27

Couldn't match expected type ‘[Char] -> [Char]’ with actual type

[Char]

The function

isValid

is applied to one argument, but its type

[Char]

has none … In the expression: isValid s In an equation for ‘parseMessage’: parseMessage s = isValid s where isValid = "Geldige string" _ = "Ongeldige string"

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