Read instances tend to be implemented such that they parse a string that looks like the source code for that type. This is usually the inverse of Show. λ> show "hello" "\"hello\"" λ> read "\"hello\"" :: String "hello" λ> read (show "hello") :: String "hello" On Fri, Dec 12, 2014 at 11:44 PM, Venu Chakravorty wrote:

Hello everyone,

I am new to Haskell and this might seem very naive, please bear with me.

======================================= Prelude> read "4" + 4 8 Prelude> (read "4" :: Int) + 4 8 Prelude> read "hello " ++ "world" "*** Exception: Prelude.read: no parse Prelude> (read "hello" :: String) ++ " world" "*** Exception: Prelude.read: no parse =======================================

Could someone please explain why the last two statements don't work?

My understanding is that "read" has a type of "read :: (Read a) => String -> a". So, "read "hello" " should give me an instance of type "Read" to which I am appending the string "world" (just like the first 2 cases where I get an instance of "Read" ("Int" in this case) to which I am adding another "Int" and I get a "Num" which is then displayed). I expected to see "hello world" as the output.

Is it that the type "String" is not an instance of type class "Read"? Please tell me what I am missing here.

Regards, Venu Chakravorty. _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners