May 2013 - Beginners - Haskell.org

Read file from sys
by Libor Wagner 11 May '13

11 May '13

Hi, I have small code snippet to read content of a file, which should grow into simple tool to read temperature from sensor on Raspberry Pi: module Main where import System.Environment main :: IO () main = do [f] <- getArgs s <- readFile f putStr s When this is called with the test file I have copied from the real location it works fine: $ cat /sys/bus/w1/devices/28-000004e46240/w1_slave >test.txt $ ./ds18b20 test.txt 53 01 4b 46 7f ff 0d 10 e9 : crc=e9 YES 53 01 4b 46 7f ff 0d 10 e9 t=21187 How ever when called with the real location nothing happens: $ ./ds18b20 /sys/bus/w1/devices/28-000004e46240/w1_slave Can someone help? I have thought that the file is not properly ended so I have tried to take just first line but that doest work either. Thanks Libor

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cabal: library and executable
by harry 09 May '13

09 May '13

I've written a library, and a simple program to demonstrate it. I can write the cabal file for the library, but I can't get the executable section right. How can I tell it to use the library defined herein?

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Type definition doesn't match with number of arguments
by Emanuel Koczwara 08 May '13

08 May '13

Hi, Please help me understund this code: http://hackage.haskell.org/packages/archive/daemons/0.1.2/doc/html/src/Cont… The type of commandSender is: (Serialize a, Serialize b) => a -> Handler (Maybe b) But definition looks like this: commandSender command reader writer = do ... As I understund, command has type a and this function returns Handler (Maybe b). reader and writer doesn't fit here. What's going on? Emanuel

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Please help me to understand: ($ 3)
by Costello, Roger L. 04 May '13

04 May '13

Hi Folks, The type signature of ($) is: ($) :: (a -> b) -> a -> b That tells me that ($) is a function that takes two arguments: 1. A function which maps values of type "a" to values of type "b" 2. A value of type "a" and it returns a value of type "b". Okay, I can understand that. So, I created some examples: ($) odd 3 -- returns True odd $ 3 -- returns True Good. But then I saw this in an article: ($ 3) odd What does ($ 3) mean? I thought the first argument to ($) is a function? I checked the type of ($ 3) and it is: ($ 3) :: Num a => (a -> b) -> b I don't understand that. How did that happen? Why can I take a second argument and wrap it in parentheses with ($) and then that second argument pops out and becomes the argument to a function? I decided to see if other functions behaved similarly. Here is the type signature for the "map" function: map :: (a -> b) -> [a] -> [b] That looks very similar to the type signature for ($). So, I reasoned, I should be able to do the same kind of thing: let list=[1,2,3] (map list) odd But that fails. Why? Why does that fail whereas a very similar looking form succeeds when ($) is used? /Roger

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How best to do this?
by emacstheviking 02 May '13

02 May '13

I have built a library for using the Hexwax expandIO-USB chip and I have now got some code to drive a stepper motor: doOption :: HWHandle -> Flag -> IO () doOption dev (Backward n) = do putStrLn $ "> STEP BACKWARD " ++ (show n) let x = [ stepBit b | b <- [3..0]] return () where stepBit p b = setBit p b 0 >> setBit p b 1 where setBit p b s = HW.setPortBit dev p b s >> stepDelay It feels a little smelly... if I don't assign the result of the list comprehension to "x" the return type blows at me hence the return(), to get past the door steward type checker... so what would be the "ideal" way to just execute my LC to step the bits and then return. I thought of using "liftIO" but I think that would be "backwards" here if you catch my drift. I am already in the IO monad as it is, so that has confused me because liftIO lifts something "into" the IO monad and I am already there... this project is making me work harder at my Haskell and I think that once I've completed it I will be much further advanced with it... So, if anybody cares to suggest more elegant and Haskell like ways to re-factor the above code I am more than happy to know about it! All the best, Sean.

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Re: [Haskell-beginners] Broken packages after upgrading to Ubuntu 13.04
by Johannes Engels 02 May '13

02 May '13

Thank you for all your valuable replies! The last days I experimented, mostly with "cabal install world"; although this terminated regularly, my broken packages were not repaired. I guess this was because many of them are not "user" (local), which seems to be a prerequisite according to Daniel Fischer's detailed advice here: http://stackoverflow.com/questions/7961604/fixing-issues-noted-by-ghc-pkg-c… I therefore decided to reinstall GHC and all the packages completely, what should not be too hard in my case. Unfortunately, it seems that there is no Haskell platform available yet which is compatible with ghc-7.6.2, the current ghc version in the Ubuntu 13.04 repo. So I shall wait for the upcoming release of the Haskell platform in May. I guess I should study Albert Lai's explanations in http://www.vex.net/~trebla/haskell/sicp.xhtml about cabalized packages in order to avoid similar flops in the future. (I just don't find this as exciting as learning about functors, monads, combinators ...) Thanks again for your suggestions! Johannes

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two System.IO.Strict modules
by Hong Yang 02 May '13

02 May '13

There are two modules called the same name: System.IO.Strict, documented respectively at http://hackage.haskell.org/packages/archive/strict-io/0.2.1/doc/html/System… http://hackage.haskell.org/package/strict-0.3.2 My questions are: 1) Will these two modules coexist in the same version of GHC? Or put it another way, both modules' .tar.gz files have System/IO/Strict.hs inside, will they overlap in physical disk? From where can I tell which locations they will be installed to? 2) If yes, which module will "import System.IO.Strict" actually use?

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