February 2013 - Beginners

"downcasting"
by Emmanuel Touzery 29 Mar '13

29 Mar '13

Hello, using the Text.Json library (and I'm reading I should be using Aeson instead... I'll migrate to it later), I'm a bit frustrated. At one point I'm getting a JSValue object, which could be anything, including a JSObject. But in this case I know I'm looking at a JSObject. I don't mind if Haskell crashes if that's not the case (for now), either it's a JSObject or the JSON I'm parsing is invalid. I hoped to do that conversion using pattern matching by giving non-exhaustive pattern match, considering only cases where the JSValue is in fact a JSONObject but somehow I can't make it... getObject :: JSValue -> JSObject JSValue getObject x@(JSObject _) = x --> doesn't compile the function returns a JSValue, I somehow wanted to point out to the compiler that if it gets there it's in fact a JSObject. getObject :: JSValue -> JSObject JSValue getObject (JSObject x) = JSObject x -> same as with the other attempt.. the compiler is saying the function is returning a JSValue: Couldn't match expected type `JSObject JSValue' with actual type `JSValue' In the return type of a call of `JSObject' In the expression: JSObject x In an equation for `getObject': getObject (JSObject x) = JSObject x I think I'm missing an obvious pattern in Haskell here, probably something I've already used elsewhere too, but somehow I don't make it. Otherwise answers saying that if I need to make such a "cast" then I'm using that library the wrong way are welcome (together to maybe a link to a tutorial explaining it the right way), but by curiosity I would still be interested in how to actually do that "cast" in this case. Thank you! Emmanuel

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Re: [Haskell-beginners] System.USB.writeInterrupt -- confused by error message from type system
by emacstheviking 06 Mar '13

06 Mar '13

Karol, Alexander, Thanks for your feedback... I am still a little confused as I shall explain... first of all let's look at the prototype for 'writeInterrupt', writeInterrupt :: DeviceHandle -> EndpointAddress -> WriteAction To me, that reads a "takes a device handle and an endpoint address and returns a WriteAction", and to quote the WriteAction help text verbatim so there is no confusion: type WriteAction = ByteString -> Timeout -> IO (Size, Status)Source Handy type synonym for write transfers. "A WriteAction is a function which takes a ByteString to write and a Timeout. The function returns an IO action which, when exectued(sic), returns the number of bytes that were actually written paired with a Status flag which indicates whether the transfer Completed or TimedOut." Now let's move to my original code and the 'right' code... action <- writeInterrupt handle endPoint let action = writeInterrupt handle endPoint If I understand things correctly up to this point, my mistake was being too eager in using "<-", my mind at that point was obviously confusing the return value from WriteAction with the return type of writeInterrupt and I can see now that what I should have done was use "let" which captures the WriteAction that is returned which can be executed with the payload and the timeout on the next line: (size, status) <- action payload 1000 On this line, the use of "<-" is what is required in order to cause the promised IO action to perform its duties and return me the tuple of data sent and status returned from the USB inner workings. However, we now come to the new type checker error, and this one has me floored right now. Time and time again I find myself beating my head against a wall and tearing my hair out trying to understand the thousand-and-one variations on strings in Haskell! I even tried the "string-conversions" (convertString) package but decided to battle it out instead... First the new code as edited in response to Karol: testBoard :: Device -> DeviceHandle -> IO () testBoard dev handle = do putStrLn $ "Inspecting device: \"" ++ (show dev) ++ "\"" -- write 0x00 0x00 0x00 0x00, get back same...we need to pad the -- packet out to 64 bytes for a full-speed device... should probably -- get this (64) from the device configuration / description record -- for maximum portability! let payload = BS.replicate 64 '\0' let endPoint = EndpointAddress 0 Out let action = writeInterrupt handle endPoint (size, status) <- action payload 1000 return () And the new error: usb1.hs:64:28: Couldn't match expected type `bytestring-0.9.2.1:Data.ByteString.Internal.ByteString' with actual type `ByteString' In the first argument of `action', namely `payload' In a stmt of a 'do' block: (size, status) <- action payload 1000 In the expression: do { putStrLn $ "Inspecting device: \"" ++ (show dev) ++ "\""; let payload = BS.replicate 64 '\NUL'; let endPoint = EndpointAddress 0 Out; let action = writeInterrupt handle endPoint; .... } Where and why does it think that "Data.ByteString.Internal.ByteString" is the type of the first parameter to "action" which is quite clearly stated as being "ByteString" ??? I know that "String", the native type is 4-bytes and that ByteString (Strict) and ByteString (Lazy) are both 8-bit, which is great, and I understand that the strict version (at least to me) feels like the rightmatch to be using for data buffers for a USB transfer but why oh why oh why can't I understand why the type checker picked up "internal" somewhere along the way? In the source code for WriteAction we have this: type WriteAction = B.ByteString → Timeout → IO (Size, Status) and at the top of the that source file: -- from bytestring: import qualified Data.ByteString as B ( ByteString, packCStringLen, drop, length ) import qualified Data.ByteString.Internal as BI ( createAndTrim, createAndTrim' ) import qualified Data.ByteString.Unsafe as BU ( unsafeUseAsCStringLen ) So why is it trying to be "internal"! I have tried not to be lazy, I have read everything and looked everywhere before posting again. If it had said: type WriteAction = BI.ByteString → Timeout → IO (Size, Status) I would have understood but it doesn't does it ?! Can somebody explain for me so I can just get on and write my killer USB application please! LOL :) Thanks, Sean. On 27 February 2013 12:07, Karol Samborski <edv.karol(a)gmail.com> wrote: > Hi Sean, > > I think that your function for testing board should look like this: > > testBoard :: Device -> DeviceHandle -> IO () > testBoard dev handle = do > putStrLn $ "Inspecting device: \"" ++ (show dev) ++ "\"\n" > -- write 0x00 0x00 0x00 0x00, get back same... > let payload = pack "\x00\x00\x00\x00" > let endPoint = EndpointAddress 0 Out > let action = writeInterrupt handle endPoint > (size, status) <- action payload 1000 > return () > > You need to use let because writeInterrupt returns (Timeout -> > ByteString -> IO (Size, Bool)) instead of IO (Timeout -> ByteString -> > IO (Size, Bool)) > > Karol > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners >

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unix-2.6.0.1 fails to build on the current haskell platform on the mac
by Miguel Negrao 27 Feb '13

27 Feb '13

Hi I’m trying to install iCalendar which depends on some package which depends on unix-2.6.0.1. unix-2.6.0.1 does not build on the mac with ghc 7.4. The issue has been reported at http://hackage.haskell.org/trac/ghc/ticket/7359, but I don’t think it’s fixed yet. It appears unix-2.6.0.1 works fine with ghc 7.6.1 but the current haskell platform comes with 7.4. What are my options ? Should I install ghc 7.6 ? If I do that what else do I have to install/update to make everything functional again ? best, Miguel http://www.friendlyvirus.org/miguelnegrao/

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System.USB.writeInterrupt -- confused by error message from type system
by emacstheviking 27 Feb '13

27 Feb '13

Hello, I am building a USB gadget with a HexWax ExpandIO-USB chip, here: http://hexwax.com/Products/expandIO-USB/ The device is working fine (lsusb shows it) and so I thought that in the name of continued training and doing things the hard way, that I would use Haskell to talk to it! I have only ever done USB with 'C' in the past and I am now stuck fighting the type checker. Here is my full code so far, it works in so far as it finds the device, opens and closes a connection and prints out some stuff. I then wanted to send four NUL bytes which elicits the same as a response according to the documentation but I cannot get it to compile. Apologies for it in advance, it's the journey you know, not the arrival, anyway, here it is... module Main where import Data.ByteString.Char8 as BS hiding (putStrLn) import Data.Word (Word16) import Data.Vector as V (filterM, null, (!)) --as V hiding ((++)) import System.USB import System.Environment -- hexwax expandIO-USB default code: "0b40:0132" main :: IO () main = do findHexWax (0xb40,0x132) >>= \hw -> case hw of Just dev -> hexwaxGo dev Nothing -> putStrLn "HexWax device not found" findHexWax :: (Word16,Word16) -> IO (Maybe Device) findHexWax (cVendor, cProd) = do usbCtx <- newCtx usbDevs <- getDevices usbCtx setDebug usbCtx PrintWarnings boards <- V.filterM (isTarget (cVendor, cProd)) usbDevs case V.null boards of True -> return Nothing False -> return $ Just $ boards!0 where isTarget :: (Word16, Word16) -> Device -> IO Bool isTarget (cVendor, cProd) dev = do info <- getDeviceDesc dev let vid = deviceVendorId info let vpr = deviceProductId info return $ ((vid, vpr) == (cVendor, cProd)) hexwaxGo :: Device -> IO () hexwaxGo dev = withDeviceHandle dev (testBoard dev) testBoard :: Device -> DeviceHandle -> IO () testBoard dev handle = do putStrLn $ "Inspecting device: \"" ++ (show dev) ++ "\"\n" -- write 0x00 0x00 0x00 0x00, get back same... let payload = pack "\x00\x00\x00\x00" let endPoint = EndpointAddress 0 Out action <- writeInterrupt handle endPoint (size, status) <- action payload 1000 return () And the error message: Prelude> :r [1 of 1] Compiling Main ( usb1.hs, interpreted ) usb1.hs:60:13: Couldn't match expected type `IO t0' with actual type `WriteAction' In the return type of a call of `writeInterrupt' Probable cause: `writeInterrupt' is applied to too few arguments In a stmt of a 'do' block: action <- writeInterrupt handle endPoint In the expression: do { putStrLn $ "Inspecting device: \"" ++ (show dev) ++ "\""; let payload = pack "\NUL\NUL\NUL\NUL"; let endPoint = EndpointAddress 0 Out; action <- writeInterrupt handle endPoint; .... } Failed, modules loaded: none. I have spent a few hours trying this and that, my Haskell is improving but not much good if I can't compile my program now is it! So, can anybody point out the error of my ways here please as I just can't see it! Thanks, Sean Charles. PS: Code style comments, alternative ways of making it more concise (but still readable and understandable mind you) are also welcome.

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Use a list as data for a data type?
by Bryce Verdier 24 Feb '13

24 Feb '13

Lets say I have a data type like so: data IpAddress = IpAddress {$ o1 :: !Int,$ o2 :: !Int,$ o3 :: !Int,$ o4 :: !Int$ } deriving (Show)$ which you can create by doing: IpAddress 192 168 1 1 Is there a way to do the same thing with the input data in a list? Like: IpAddress [192,168,1,1] or using some operator (or combination of operators):Like (I'm guessing (<????>) isn't in use): IpAddress <????> [192,168,1,1] Thanks in advance, Bryce

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Re: [Haskell-beginners] Beginners Digest, Vol 56, Issue 33
by xiao Ling 24 Feb '13

24 Feb '13

Hi All: How do you define a function of signature h :: M Int -> M Int -> M Int so that h ( M x ) ( M y ) = M ( x + y ), but without unwrapping the value from the monad? This question is from the article "Trivial Monad" found at http://blog.sigfpe.com/2007/04/trivial-monad.html. The provided answer is h x y = x >>= (\x -> g x y) or equivalently ( in context of the article ) h :: M Int -> M Int -> M Int h x y = bind ( \x-> g x y ) x where g is g :: Int -> W Int -> W Int g x y = y >>= (return . (+x)) for the monad: data M a = M a deriving Show Now I am a little confused, how can you put x in g if it takes an Int as first parameter but x is M Int? Thanks!

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How to extend fuinctions?
by Heinrich Ody 22 Feb '13

22 Feb '13

Hi, I have a function Delta and two sets S and A. Delta maps some elements from (S x A) to S. How can I extend Delta to some Delta' such that it assigns to every pair in (S x A) that Delta is not defined for a value not occuring in S? Something like: Delta' = if (Delta s a) /= _Undefined_ then Delta s a else if () n where n is the new value. However Delta' should not assign values to pairs not in (S x A). Note: (S x A) is meant to be the set-crossprroduct. Background: I want to learn Haskell by building a small library for Finite State Automata, Büchi Automata etc. I need this extension of Delta for making automata complete. Regards and thanks for reading, Heinrich

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How do I declare a set?
by Heinrich Ody 22 Feb '13

22 Feb '13

Hi, with [1,2] I can declare a list. Is there a similar notation for sets? Currently the only way I know is 'Set.fromList [1,2]' which is unhandy... Regards, Heinrich

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Is my understanding of ">>=" correct?
by Martin Drautzburg 22 Feb '13

22 Feb '13

Hello all, It took me quite some time to understand >>=. What confused me, was that the second operand is a function, not just a Monad. I believe I am beginning to see the light. First there is no way to sequence operations by just writing one after the other. However we can chain operations by passing the output of one operation to the next. But this does not not allow capturing the intermediate results. With >>= the output of one operation gets magically bound to the parameter of the function and remains in scope all the way through. I was confused by the signature m a -> (a -> mb) -> mb because it looks like there is (as second parameter) a function which returns a Monad. But I doubt I ever saw such a function, in the sense that it does something intelligent. It is typically a function that does not do a whole lot with its parameter, but just returns another Monad, such that its parameter is bound to the result of the previous expression. But it could be done: do x <- SomeMonad y <- someCleverlyConstructedMonad x If I want to see the function of the >>= signature I have to read the do block right->left (following the arrow) and then diagonally to the lower right. The diagonal steps reveal the function. In the above example we have \x -> someCleverlyConstructedMonad but many times I just see do x <- SomeMonad y <- someOtherMonad and the function does not do anything intelligent but just binds x and enforces the sequence of operations. is this about right? -- Martin

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Stack overflow folding a map
by tcollin371 21 Feb '13

21 Feb '13

I've been playing around with Haskell on and off for a while, and am getting a "Stack space overflow" that I can't seem to correct. I have a map containing about 2 million items. Actually, it's a map keyed off of an Int where the element is another map keyed off of a string that has three Ints as its data. The number of top level maps is pretty small (< 50), and the maps held by them hold a lot of entries. I want to walk through all of the entries and gather some statistics, but I keep getting the stack overflow. I worked down my test to an inner and outer fold that just counts the entries, and I still get the stack overflow. I added 'seq' calls to make sure that the count isn't building up a bunch of undone operations, and I'm using foldlWithKey, and have also tried foldlWithKey' with the same result. I get the count in my main function and immediately print it out. Here is the relevant code: outputIfNotFilteredCount1 :: Int -> KeyMap -> PuzzleMapRecord -> Int outputIfNotFilteredCount1 inCount key entry = seq inCount (inCount + 1) outputDatabaseCount1 :: Int -> Int -> InnerDB -> Int outputDatabaseCount1 inCount smndCount innerDB = seq inCount (inCount + outCount) where outCount = Map.foldlWithKey' outputIfNotFilteredCount1 0 innerDB main = do ... let count = Map.foldlWithKey' outputDatabaseCount1 0 resultDB putStrLn $ "Database entries:" ++ (show count) Any thoughts on why this is running out of stack space? -Truman

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