Network.Socket.ByteString.recv uses the strict ByteString from
Data.ByteString, not the lazy one from Data.ByteString.Lazy. So you
want the `unpack` from Data.ByteString.Char8, rather than
Data.ByteString.Lazy.Char8.
I never remember which functions return strict or lazy ByteString. I
find the easiest way to check is to open the online docs and see where
the `ByteString` link points:
https://hackage.haskell.org/package/network-2.7.0.0/docs/Network-Socket-Byte...
points to:
https://hackage.haskell.org/package/bytestring-0.10.8.2/docs/Data-ByteString...
hope this helps,
bergey
On 2018-05-20 at 07:52, Dinesh Amerasekara
Hi,
I am unable to compile the below code.
import Network.Socket hiding(recv) import Network.Socket.ByteString as S (recv) import qualified Data.ByteString.Lazy.Char8 as Char8
getMessage :: Socket -> IO String getMessage sock = Char8.unpack <$> S.recv sock 8888
It gives the below error.
Couldn't match type ‘Data.ByteString.Internal.ByteString’ with ‘ByteString’ NB: ‘ByteString’ is defined in ‘Data.ByteString.Lazy.Internal’ ‘Data.ByteString.Internal.ByteString’ is defined in ‘Data.ByteString.Internal’ Expected type: IO ByteString Actual type: IO Data.ByteString.Internal.ByteString
In the second argument of ‘(<$>)’, namely ‘recv sock 8888’ In the expression: unpack <$> recv sock 8888 In an equation for ‘getMsg’: getMsg sock = unpack <$> recv sock 8888
Can somebody tell me how I can return the IO String using Network.Socket.ByteString.recv?
Best Regards, Dinesh.
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