On Sunday 31 July 2011, 23:24:13, Alexey G wrote:
Hello. I have some problem with catching error.
My function try to read value from string and if it's fails - return default (d).
With this:
import Control.Exception readIOWith :: Read a => a -> String -> IO a readIOWith d x = catch tryRead errHandler
where errHandler :: SomeException -> IO a
There's your problem. The type variable 'a' in the local signature is completely unrelated to the type variable 'a' in the top-level signature (type variables have an implicit forall in Haskell), so the local signature says "errHandler can return `IO anything'", but it can of course only return `IO (Type of d)'. If you remove the signature, it should compile. (Another possibility is to bring the type variable into scope, that would require the ScopedTypeVariables language extension and an explicit forall.) However, this shouldn't need any IO at all, import Data.Char -- for isSpace readWithDefault :: Read a => a -> String -> a readWithDefault d s = case reads s of [(result,remaining)] | all isSpace remaining -> result _ -> d If you really need the IO return type: readIOWith d = return . readWithDefault d
errHandler _ = return d tryRead = readIO x
I have error:
Could not deduce (a ~ a1) from the context (Read a) bound by the type signature for readIOWith :: Read a => a -> String -> IO a at BWClub/Common/Helpers/Packets.hs:(19,1)-(22,33) `a' is a rigid type variable bound by the type signature for readIOWith :: Read a => a -> String -> IO a at BWClub/Common/Helpers/Packets.hs:19:1 `a1' is a rigid type variable bound by the type signature for errHandler :: SomeException -> IO a1 at BWClub/Common/Helpers/Packets.hs:21:11 In the first argument of `return', namely `d' In the expression: return d In an equation for `errHandler': errHandler _ = return d
But this code works:
readIOWith d x = catch tryRead (\e -> print (e :: SomeException) >> return
d)
where tryRead = readIO x
Here, there's no signature introducing a fresh type variable.
Sorry for my english.