main :: IO ()
main = do
let a = ("teeeeeeeeeeeeest","teeeeeeeeeeeest")
b <- app a
print b
app expects (bytestring,bytestring) not ([char],[char])
how can I convert it?
You can convert Strings to ByteStrings with Data.ByteString.Char8.pack (or the lazy ByteString version thereof) if your String contains only ASCII values or you are interested only in the last eight bits of each Char,
import qualified Data.ByteString.Char8 as C
main :: IO ()
main = do
let a = ("teeeeeeeeeeeeest","teeeeeeeeeeeest")
b <- app $ (\(x,y) -> (C.pack x, C.pack y)) a
print b
If your String contains non-ASCII Chars and you are interested in more than only the last eight bits, you will need some other encoding, like Data.ByteString.UTF8.fromString.
You could try:
import qualified Data.ByteString.Char8 as B --to prevent name clash with Prelude
B.pack "Hello, world"
A lot of useful functions can be found here:
http://www.haskell.org/ghc/docs/latest/html/libraries/bytestring/Data-ByteString-Char8.html
you could also use Data.ByteString.Lazy.Char8
for lazy bytestrings
http://hackage.haskell.org/packages/archive/bytestring/latest/doc/html/Data-ByteString-Lazy-Char8.html#v:pack
Related
I'm trying to pass a random string (which happens to be a number) "4176730.5" to SHA in Haskell to get a larger random string like "2d711642b726b04401627ca9fbac32f5c8530fb1903cc4db02258717921a4881".
I have this code to generate a random number and cast it to a string
num <- randomIO :: IO Float
let x = C.pack (show (num*10000000))
print x
but when I pass it to SHA with
let a = sha256 x
I get the error
Couldn't match expected type ‘Data.ByteString.Lazy.Internal.ByteString’
with actual type ‘C.ByteString’
I've tried casting my number to C.ByteString, but I think there are two types of Bytestring, according to the Haskell compiler.
The full code is:
import Data.Digest.Pure.SHA
import System.Random
import qualified Data.ByteString.Char8 as C
main :: IO ()
main = do
num <- randomIO :: IO Float
let x = C.pack (show (num*10000000))
print x
let a = sha256 x
b = hmacSha256 "key" "some test message"
mapM_ print [showDigest a, showDigest b]
Seeing as how there are apparently two types of Bytestring, and I'm casting to the wrong one, how do I cast my random string correctly?
Further to #Cubic's answer below if I replace
import qualified Data.ByteString.Char8 as C
with
import qualified Data.ByteString.Lazy as C
I just get these errors
Couldn't match type ‘Char’ with ‘GHC.Word.Word8’
Expected type: [GHC.Word.Word8]
Actual type: String
and
Couldn't match expected type ‘C.ByteString’
with actual type ‘[Char]’
The issue is that a ByteString is a sequence of bytes, while a String is a sequence of chars. There are many ways to turn chars into bytes, so you need to specify which encoding you want. Most likely, you want an ASCII or UTF8 encoding. If so, you can use this solution below, which converts strings into "UTF8 bytes" as needed.
import Data.Digest.Pure.SHA
import System.Random
import qualified Data.ByteString.Lazy as C
import qualified Data.ByteString.Lazy.UTF8 as U
main :: IO ()
main = do
num <- randomIO :: IO Float
let x = U.fromString (show (num*10000000))
print x
let a = sha256 x
b = hmacSha256 (U.fromString "key") (U.fromString "some test message")
mapM_ print [showDigest a, showDigest b]
You need Data.ByteString.Lazy, not Data.ByteString.Char8.
In general, you almost never want Data.ByteString.Char8.
Just use the lazy bytestrings as #leftaroundabout mentioned. Your try didn't work because you want to pack from Strings, so you need to import the .Char8 module to achieve that:
import Data.ByteString.Lazy.Char8 as C
I'm testing some HTTP requests in haskell and have the below methods:
import qualified Data.ByteString.Lazy as LAZ
import Language.Haskell.TH.Ppr
import System.IO
import Data.Word (Word8)
request :: IO LAZ.ByteString
request = do
response <- simpleHttp "https://www.url.com"
return (response)
exampleFunctionOne:: IO LAZ.ByteString -> IO LAZ.ByteString
exampleFunctionOne bytes = do
html <- bytes
let bytesToChars = bytesToString $ LAZ.unpack html
let x = exampleFunctionTwo bytesToChars
bytes
exampleFunctionTwo :: [Char] -> [Char]
exampleFunctionTwo chars = --Do stuff...
main = do
exampleFunctionOe $ request
My questions are:
Is there a more straight forward way to convert the ByteString to [Char]? Currently I've having to convert to perform (ByteString -> Word8) and then (Word8 -> Char)
Am I correct in saying the 'return ()' statement in my request function is simply re-applying the monad context (in this case IO) to the value I've extracted (response <- simpleHttp)? Or does it have an additional purpose?
To answer your first question, note that there's a different "unpack" in Data.ByteString.Lazy.Char8 with the signature you want:
unpack :: ByteString -> String
It's not unusual for people to import both modules:
import qualified Data.ByteString.Lazy as B
import qualified Data.ByteString.Lazy.Char8 as C
and mix and match functions from each.
To answer your second question, yes that's more or less it. For example:
redund = do x <- getLine
y <- return x
z <- return y
u <- return z
return u
is all equivalent to redund = getLine with a bunch of re-wrapping and extracting of pure values into an out of an IO monad.
While writing a deserialiser for a large (<bloblength><blob>)* encoded binary file I got stuck with the various Haskell produce-transform-consume libraries. So far I'm aware of four streaming libraries:
Data.Conduit: Widely used, has very careful resource management
Pipes: Similar to conduit (Haskell Cast #6 nicely reveals the differences between conduit and pipes)
Data.Binary.Get: Offers useful functions such as getWord32be, but the streaming example is awkward
System.IO.Streams: Seems to be the easiest one to use
Here's a stripped down example of where things go wrong when I try to do Word32 streaming with conduit. A slightly more realistic example would first read a Word32 that determines the blob length and then yield a lazy ByteString of that length (which is then deserialised further).
But here I just try to extract Word32's in streaming fashion from a binary file:
module Main where
-- build-depends: bytestring, conduit, conduit-extra, resourcet, binary
import Control.Monad.Trans.Resource (MonadResource, runResourceT)
import qualified Data.Binary.Get as G
import qualified Data.ByteString as BS
import qualified Data.ByteString.Char8 as C
import qualified Data.ByteString.Lazy as BL
import Data.Conduit
import qualified Data.Conduit.Binary as CB
import qualified Data.Conduit.List as CL
import Data.Word (Word32)
import System.Environment (getArgs)
-- gets a Word32 from a ByteString.
getWord32 :: C.ByteString -> Word32
getWord32 bs = do
G.runGet G.getWord32be $ BL.fromStrict bs
-- should read BytesString and return Word32
transform :: (Monad m, MonadResource m) => Conduit BS.ByteString m Word32
transform = do
mbs <- await
case mbs of
Just bs -> do
case C.null bs of
False -> do
yield $ getWord32 bs
leftover $ BS.drop 4 bs
transform
True -> return ()
Nothing -> return ()
main :: IO ()
main = do
filename <- fmap (!!0) getArgs -- should check length getArgs
result <- runResourceT $ (CB.sourceFile filename) $$ transform =$ CL.consume
print $ length result -- is always 8188 for files larger than 32752 bytes
The output of the program is just the number of Word32's that were read. It turns out the stream terminates after reading the first chunk (about 32KiB). For some reason mbs is never Nothing, so I must check null bs which stops the stream when the chunk is consumed. Clearly, my conduit transform is faulty. I see two routes to a solution:
The await doesn't want to go to the second chunk of the ByteStream, so is there another function that pulls the next chunk? In examples I've seen (e.g. Conduit 101) this is not how it's done
This is just the wrong way to set up transform.
How is this done properly? Is this the right way to go? (Performance does matter.)
Update: Here's a BAD way to do it using Systems.IO.Streams:
module Main where
import Data.Word (Word32)
import System.Environment (getArgs)
import System.IO (IOMode (ReadMode), openFile)
import qualified System.IO.Streams as S
import System.IO.Streams.Binary (binaryInputStream)
import System.IO.Streams.List (outputToList)
main :: IO ()
main = do
filename : _ <- getArgs
h <- openFile filename ReadMode
s <- S.handleToInputStream h
i <- binaryInputStream s :: IO (S.InputStream Word32)
r <- outputToList $ S.connect i
print $ last r
'Bad' means: Very demanding in time and space, does not handle Decode exception.
Your immediate problem is caused by how you are using leftover. That function is used to "Provide a single piece of leftover input to be consumed by the next component in the current monadic binding", and so when you give it bs before looping with transform you are effectively throwing away the rest of the bytestring (i.e. what is after bs).
A correct solution based on your code would use the incremental input interface of Data.Binary.Get to replace your yield/leftover combination with something that consumes each chunk fully. A more pragmatic approach, though, is using the binary-conduit package, which provides that in the shape of conduitGet (its source gives a good idea of what a "manual" implementation would look like):
import Data.Conduit.Serialization.Binary
-- etc.
transform :: (Monad m, MonadResource m) => Conduit BS.ByteString m Word32
transform = conduitGet G.getWord32be
One caveat is that this will throw a parse error if the total number of bytes is not a multiple of 4 (i.e. the last Word32 is incomplete). In the unlikely case of that not being what you want, a lazy way out would be simply using \bs -> C.take (4 * truncate (C.length bs / 4)) bs on the input bytestring.
With pipes (and pipes-group and pipes-bytestring) the demo problem reduces to combinators. First we resolve the incoming undifferentiated byte stream into little 4 byte chunks:
chunksOfStrict :: (Monad m) => Int -> Producer ByteString m r -> Producer ByteString m r
chunksOfStrict n = folds mappend mempty id . view (Bytes.chunksOf n)
then we map these to Word32s and (here) count them.
main :: IO ()
main = do
filename:_ <- getArgs
IO.withFile filename IO.ReadMode $ \h -> do
n <- P.length $ chunksOfStrict 4 (Bytes.fromHandle h) >-> P.map getWord32
print n
This will fail if we have less than 4 bytes or otherwise fail to parse but we can as well map with
getMaybeWord32 :: ByteString -> Maybe Word32
getMaybeWord32 bs = case G.runGetOrFail G.getWord32be $ BL.fromStrict bs of
Left r -> Nothing
Right (_, off, w32) -> Just w32
The following program will then print the parses for the valid 4 byte sequences
main :: IO ()
main = do
filename:_ <- getArgs
IO.withFile filename IO.ReadMode $ \h -> do
runEffect $ chunksOfStrict 4 (Bytes.fromHandle h)
>-> P.map getMaybeWord32
>-> P.concat -- here `concat` eliminates maybes
>-> P.print
There are other ways of dealing with failed parses, of course.
Here, though, is something closer to the program you asked for. It takes a four byte segment from a byte stream (Producer ByteString m r) and reads it as a Word32 if it is long enough; it then takes that many of the incoming bytes and accumulates them into a lazy bytestring, yielding it. It just repeats this until it runs out of bytes. In main below, I print each yielded lazy bytestring that is produced:
module Main (main) where
import Pipes
import qualified Pipes.Prelude as P
import Pipes.Group (folds)
import qualified Pipes.ByteString as Bytes ( splitAt, fromHandle, chunksOf )
import Control.Lens ( view ) -- or Lens.Simple (view) -- or Lens.Micro ((.^))
import qualified System.IO as IO ( IOMode(ReadMode), withFile )
import qualified Data.Binary.Get as G ( runGet, getWord32be )
import Data.ByteString ( ByteString )
import qualified Data.ByteString.Lazy.Char8 as BL
import System.Environment ( getArgs )
splitLazy :: (Monad m, Integral n) =>
n -> Producer ByteString m r -> m (BL.ByteString, Producer ByteString m r)
splitLazy n bs = do
(bss, rest) <- P.toListM' $ view (Bytes.splitAt n) bs
return (BL.fromChunks bss, rest)
measureChunks :: Monad m => Producer ByteString m r -> Producer BL.ByteString m r
measureChunks bs = do
(lbs, rest) <- lift $ splitLazy 4 bs
if BL.length lbs /= 4
then rest >-> P.drain -- in fact it will be empty
else do
let w32 = G.runGet G.getWord32be lbs
(lbs', rest') <- lift $ splitLazy w32 bs
yield lbs
measureChunks rest
main :: IO ()
main = do
filename:_ <- getArgs
IO.withFile filename IO.ReadMode $ \h -> do
runEffect $ measureChunks (Bytes.fromHandle h) >-> P.print
This is again crude in that it uses runGet not runGetOrFail, but this is easily repaired. The pipes standard procedure would be to stop the stream transformation on a failed parse and return the unparsed bytestream.
If you were anticipating that the Word32s were for large numbers, so that you did not want to accumulate the corresponding stream of bytes as a lazy bytestring, but say write them to different files without accumulating, we could change the program pretty easily to do that. This would require a sophisticated use of conduit but is the preferred approach with pipes and streaming.
Here's a relatively straightforward solution that I want to throw into the ring. It's a repeated use of splitAt wrapped into a State monad that gives an interface identical to (a subset of) Data.Binary.Get. The resulting [ByteString] is obtained in main with a whileJust over getBlob.
module Main (main) where
import Control.Monad.Loops
import Control.Monad.State
import qualified Data.Binary.Get as G (getWord32be, runGet)
import qualified Data.ByteString.Lazy as BL
import Data.Int (Int64)
import Data.Word (Word32)
import System.Environment (getArgs)
-- this is going to mimic the Data.Binary.Get.Get Monad
type Get = State BL.ByteString
getWord32be :: Get (Maybe Word32)
getWord32be = state $ \bs -> do
let (w, rest) = BL.splitAt 4 bs
case BL.length w of
4 -> (Just w', rest) where
w' = G.runGet G.getWord32be w
_ -> (Nothing, BL.empty)
getLazyByteString :: Int64 -> Get BL.ByteString
getLazyByteString n = state $ \bs -> BL.splitAt n bs
getBlob :: Get (Maybe BL.ByteString)
getBlob = do
ml <- getWord32be
case ml of
Nothing -> return Nothing
Just l -> do
blob <- getLazyByteString (fromIntegral l :: Int64)
return $ Just blob
runGet :: Get a -> BL.ByteString -> a
runGet g bs = fst $ runState g bs
main :: IO ()
main = do
fname <- head <$> getArgs
bs <- BL.readFile fname
let ls = runGet loop bs where
loop = whileJust getBlob return
print $ length ls
There's no error handling in getBlob, but it's easy to extend. Time and space complexity is quite good, as long as the resulting list is used carefully. (The python script that creates some random data for consumption by the above is here).
Trying to read a binary file of Word8, I have the following program:
import qualified Data.Binary as B
type Chars = [B.Word8]
printChars :: Chars -> IO()
printChars cs = mapM_ print cs
main :: IO()
main = do
chars <- B.decodeFile "chars"
printChars chars
When I run it, I get an error:
$ ./test
test: too few bytes. Failed reading at byte position 241
It seems decodeFile expects an infinite list.
How can I tell it to just read as many elements as possible?
Edit:
Here was the code I was looking for: (This works with any type, not just Word8.)
import Prelude hiding ( readFile )
import Data.ByteString.Lazy ( readFile )
import Data.Binary.Get ( isEmpty, runGet )
import qualified Data.Binary as B
type Chars = [B.Word8]
printChars :: Chars -> IO()
printChars cs = mapM_ print cs
-- see http://hackage.haskell.org/package/binary-0.7.1.0/docs/Data-Binary-Get.html
-- function getTrades
getChars = do
e <- isEmpty
if e then return []
else do
c <- B.get
cs <- getChars
return (c:cs)
main :: IO()
main = do
input <- readFile "chars"
printChars $ runGet getChars input
Data.Binary is used to serialize known types in a canonical way (defined by the Binary class instance). In general it isn't intended for unstructured data.
In the case you gave us, you are attempting to deserialize the bytes in a file to an object of type [B.Word8]. If you look in the Data.Binary source code, you can see the following
instance Binary a => Binary [a] where
get = do n <- get :: Get Int
getMany n
which basically means that arrays are stored as follows
[length of array, val1, val2, ....]
so, when you applied the value to the file, it read the first Int in the file (no doubt, a very large number), then attempted to read that number of values.
If you just want to load the file as bytes, you should use Data.ByteString.getContents
I am in the midst of converting a package from using GHC.IO.Handle for networking to Network.Connection.Connection. One of the pain points is the number of places where Data.ByteString.Lazy.ByteString is being changed to Data.ByteString.Char8.ByteString.
An example function is:
import qualified Data.ByteString.Lazy as BL
-- gets the size of the frame
-- the bytestring should be at least 7 bytes long, otherwise this method will fail
peekFrameSize :: BL.ByteString -> PayloadSize
peekFrameSize = runGet f
where
f = do
void $ getWord8 -- 1 byte
void $ (get :: Get ChannelID) -- 2 bytes
get :: Get PayloadSize -- 4 bytes
How do I convert Data.ByteString.Char8.ByteString to Data.ByteString.Lazy.ByteString for use in Data.Binary.Get?
First of all, Data.ByteString.Char8 is a module, not a type. Same with Data.ByteString.Lazy. That said, and assuming the obvious, then notice that Data.ByteString.Char8.ByteString is actually a re-export of Data.ByteString.ByteString. So all you need is the fromStrict function mentioned by Mikail Glushenkov.